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1 NTNU Norwegian Institute of Science and Technology Department of Physics Quantumtheoryofmany particlesystems Lecture notes for TFY4210 Jens O. Andersen Second edition 2011 |ψ= a p |0

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Page 1: Quantumtheoryofmany particlesystemsfolk.ntnu.no/jensoa/h.pdf · 2011-05-18 · Chapter 1 Classical mechanics In introductory courses, we have been taught classical mechanics and in

1

NTNUNorwegian Institute of Science and Technology

Department of Physics

Quantum theory of many − particle systemsLecture notes for TFY4210

Jens O. Andersen

Second edition 2011

|ψ〉 = a†p|0〉

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Contents

1 Classical mechanics 6

1.1 Lagrange’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Particle in external electromagnetic fields . . . . . . . . . . . . . . . . . . . . 11

1.4 Calculus of variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.5.1 Straight line as minimal distance between points I . . . . . . . . . . . 15

1.5.2 Straight line as minimal distance between points II . . . . . . . . . . 16

1.5.3 Extremal curves on a sphere - Great circles . . . . . . . . . . . . . . . 16

1.5.4 Rigid rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5.5 Mathematical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Lorentz transformations 18

2.1 Boosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Geometry of Minkowski space . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Proper time and rate of clocks . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.1 Interpretation of boosts . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.2 D’Alambertian under boosts . . . . . . . . . . . . . . . . . . . . . . . 24

2.5.3 Rotation matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.5.4 Spacelike and timelike intervals . . . . . . . . . . . . . . . . . . . . . 24

2.5.5 Lorentz contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.5.6 Compositions of boosts . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 Classical field theory 26

3.1 Action and equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2 Conjugate momentum and Hamiltonian . . . . . . . . . . . . . . . . . . . . . 29

3.3 Symmetries and conservation laws . . . . . . . . . . . . . . . . . . . . . . . . 30

2

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CONTENTS 3

3.4 Complex Klein-Gordon equation and gauge transformations . . . . . . . . . 35

3.5 Nonrelativistic limit of fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.6.1 Nonrelativistic field theory . . . . . . . . . . . . . . . . . . . . . . . . 38

3.6.2 Dual field tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.6.3 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.6.4 Free field theory for fermions . . . . . . . . . . . . . . . . . . . . . . 39

4 Relativistic Wave equations 40

4.1 Klein-Gordon equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.1.1 Relativistic corrections . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.2 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2.1 Representation independence . . . . . . . . . . . . . . . . . . . . . . 42

4.2.2 Relativistic corrections . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.1 Operator relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.3.2 Klein-Gordon equation in external fields . . . . . . . . . . . . . . . . 46

5 Field quantization 48

5.1 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5.3 Nonrelativistic field theories . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

5.4 Note on fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.5 Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.6.1 Anharmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.6.2 Momentum operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.6.3 Nonrelativistic fermions . . . . . . . . . . . . . . . . . . . . . . . . . 61

5.6.4 Integral representation of step function . . . . . . . . . . . . . . . . . 62

5.6.5 Propagator for NR field . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.6.6 Propagator for the Dirac field . . . . . . . . . . . . . . . . . . . . . . 62

5.6.7 Finite density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.6.8 Complex propagator and Greens’ function . . . . . . . . . . . . . . . 63

6 Perturbation theory 65

6.1 Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.2 Vacuum energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

6.3 Dimensional regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

6.3.1 Vacuum energy revisited . . . . . . . . . . . . . . . . . . . . . . . . . 73

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CONTENTS 4

6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.4.1 Dimensional regularization . . . . . . . . . . . . . . . . . . . . . . . . 75

7 Dilute Bose gas 76

7.1 Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

7.2 Bogoliubov approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

7.3 Bose-Einstein condensation and spontaneous symmetry breaking . . . . . . . 84

7.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

7.4.1 Complex scalar field, symmetry breaking, and Goldstone mode . . . . 87

7.4.2 Weakly interacting Bose gas . . . . . . . . . . . . . . . . . . . . . . . 87

7.4.3 Dimensional regularization . . . . . . . . . . . . . . . . . . . . . . . . 88

8 Fermi gases 89

8.1 Perfect Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

8.2 Interacting Fermi gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

8.3 Degenerate electron gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

8.3.1 Photon propagator and the Coulomb potential . . . . . . . . . . . . . 95

8.3.2 Self-interaction of homogeneous background . . . . . . . . . . . . . . 97

8.3.3 Eletron interaction with homogeneous background . . . . . . . . . . . 98

8.3.4 Eletron-electron interaction . . . . . . . . . . . . . . . . . . . . . . . 99

9 Effective potential 102

9.1 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

9.2 Dilute Bose gas revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

9.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

9.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

10 Bose-Einstein condensation in traps 109

10.1 Hydrodynamics of BEC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

A Sum vs integral 116

B Interatomic potentials and scattering processes 118

B.1 Scattering and the Born approximation . . . . . . . . . . . . . . . . . . . . . 121

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Preface

All rights reserved by the author. No parts of this book may be reproduced, in any form orby any means without permission in writing from the author. Failure to comply with this,will result in legal action.

First edition

These lecture notes are written to cover the material taught in TFY4210 Applied quantummechanics which is advanced course for fourth-year students at NTNU. It covers classicalmechanics and classical field theory including Noether’s theorem, relativistic wave equa-tions, quantization of fields Dilute Bose and Fermi gases, Bose-Einstein condensation andBCS theory. If you find typos or have comments or suggestions, please email the author:[email protected].

Trondheim May 2010

Jens O. Andersen

Second edition

The text has been expanded and improved several places. Typos have been corrected andIf you find more typos or have comments or suggestions, please email the author: [email protected].

Trondheim May 2011

Jens O. Andersen

5

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Chapter 1

Classical mechanics

In introductory courses, we have been taught classical mechanics and in particular Newton’ssecond law. There are two other formulations of classical mechanics, which are equivalent tothe Newtonian formulation, namely the Lagrangian and Hamiltonian formulations. In theNewtonian formulation, the concept of force is essential, while in the other formulations, theconcept of energy is essential. These formulations are important in the transition to quan-tum mechanics and in statistical mechanics. We recommend the classic textbook Classical

Mechanics [1] by Goldstein for a thorough treatment.

In this chapter, we derive Lagrange and Hamilton’s equations in a somewhat sloppyfashion. We will do this by considering a simple example, but the results are valid undervery general conditions.

1.1 Lagrange’s equations

Consider a single particle moving in one dimension in a potential V (x). Newton’s secondlaw becomes

mx = −dV (x)

dx, (1.1)

where m is the mass of the particle and a dot means differentiation with respect to time t.We next define the Lagrangian L of a particle, which is the difference between its kineticenergy T and it potential energy V , i.e,

L = T − V . (1.2)

and their time derivatives qi. In our example, the Lagrangian becomes

L =1

2mx2 − V (x) . (1.3)

Note that L is a function of the coordinate x and the time derivative x. Taking the partialderivatives of L with respect to x and x, we obtain

∂L

∂x= mx , (1.4)

6

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CHAPTER 1. CLASSICAL MECHANICS 7

∂L

∂x= −dV (x)

dx. (1.5)

By differentiating the first of these equations with respect to t, we see that Newton’s equa-tion (1.1) can be written as

d

dt

(

∂L

∂x

)

=∂L

∂x. (1.6)

This is the Lagrange equation for a particle in one dimension. We have derived it for a verysimple mechanical problem, but the result can be generalized to any number of particles andany number of dimensions. If we consider a system of N particles in three dimensions, weneed 3N coordinates to specify the positions of the particles. These coordinates are calledgeneralized coordinates and are denoted by q1, q2, ..., q3N

1. The Lagrangian is then a functionof the coordinates q1, q2, ..., q3N , their time derivatives q1, q2, ..., q3N , and possibly t explicitly,i.e. L(q1, ..., q3N , t). In condensed notation, we write L = L(q, q, t). There is one equationfor each coordinate qi and Lagrange equations are

ddt

(

∂L∂qi

)

= ∂L∂qi

The Lagrange equations form a set of second-order differential equations.

If the Lagrangian is independent of one of the coordinates, qi, i.e. if

∂L

∂qi= 0 , (1.7)

then the corresponding coordinate qi is called a cyclic coordinate. Integration of Lagrange’sequation then immediately gives

∂L

∂qi= C . (1.8)

Thus a cyclic coordinate leads to a conserved quantity. Conserved quantities result fromsymmetries of the system.——————————————————————————————————

Example

We consider a single particle moving in a plane under the influence of a rotationally sym-metric potential V (r). The Lagrangian in cartesian coordinates is then given by

L =1

2m(

x2 + y2)

− V(

x2 + y2

)

. (1.9)

1We can use any coordinates, cartesian coordinates, spherical coordinates...Hence the name generalizedcoordinates.

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CHAPTER 1. CLASSICAL MECHANICS 8

Using the relation between x and y, and r and θ

x = r cos θ , (1.10)

y = r sin θ , (1.11)

differentiation with respect to time yields:

x = r cos θ − rθ sin θ , (1.12)

y = r sin θ + rθ cos θ . (1.13)

Substituting Eqs. (1.12) and (1.13) into Eq. (1.9), the Lagrangian in polar coordinates be-comes

L =1

2m(

r2 + r2θ2)

− V (r) . (1.14)

Note that the θ is a cyclic coordinate as it does not appear explicitly in the Lagrangian (1.14).The corresponding conserved quantity is

∂L

∂θ= mr2θ , (1.15)

which is nothing but the z-component of the angular momentum.——————————————————————————————————

1.2 Hamilton’s equations

Given the Lagrangian, we next define the conjugate momentum or generalized momentum pito a coordinate qi by

pi =∂L

∂qi. (1.16)

If qi is a usual cartesian coordinate, the generalized momentum is equal to the familiarlinear momentum of a particle. If qi is an angular coordinate, as in the example above, thegeneralized momentum is the angular momentum.

The Hamiltonian H is defined as

H =∑

i

piqi − L . (1.17)

In most cases, the Hamiltonian is the sum of the kinetic and potential energy H = T+V . Wenext show that H is a function of pi and qi and so can be interpreted as Legendre transformof the Lagrangian. To this end, take the differential of the above equation

dH =∑

i

pidqi +∑

i

qidpi −∑

i

∂L

∂qidqi −

i

∂L

∂qidqi (1.18)

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CHAPTER 1. CLASSICAL MECHANICS 9

Using the definition of the generalized momentum pi, we see that the first and third termcancel and we are left with

dH =∑

i

qidpi −∑

i

∂L

∂qidqi . (1.19)

Thus the differential of H is given in terms of the differentials of pi and qi, which shows thatH = H(p, q) in condensed notation. Using this fact, the differential of H becomes

dH =∑

i

∂H

∂pidpi +

i

∂H

∂qidqi . (1.20)

Comparing Eqs. (1.19) and (1.20), we obtain Hamilton’s equations

∂H∂pi

= qi

∂H∂qi

= −pi

Notice that Hamilton’s equations are first order in time, but there are twice as many ascompared to Lagrange equations. The two sets of equations represent two equivalent waysof describing the same physics.

Now reconsider the problem of a single particle moving in one dimension in a potentialV (x). The linear momentum p is given by

∂L

∂x= mx . (1.21)

This implies that x = p/m. The Hamiltonian is then

H = px− L

=p2

m− p2

2m+ V (x)

=p2

2m+ V (x)

= T + V . (1.22)

Hence the Hamiltonian is the total energy of the particle. We have shown this for a verysimple example, but again it holds very generally for systems of many particles where weare using any coordinate system we wish.——————————————————————————————————

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CHAPTER 1. CLASSICAL MECHANICS 10

Example

Reconsider the previous example. The generalized momenta pr and pθ corresponding to thepolar coordinates r and θ follow from the Lagrangian (1.14) and the definition (1.16):

pr = mr , (1.23)

pθ = mr2θ . (1.24)

The Hamiltonian H can then be written as

H = prr + pθθ − L . (1.25)

Remember that H must be expressed in terms of pr, pθ, r, and θ. It is therefore necessaryto solve Eqs. (1.23) and (1.24) with respect to r and θ in order to eliminate these quantitiesin L in favor of the p’s and the q’s. Using that r = pr/m and θ = pθ/mr

2, we obtain

H =1

m

(

p2r +

p2θ

r2

)

− 1

2m

(

p2r +

p2θ

r2

)

+ V (r)

=1

2m

(

p2r +

p2θ

r2

)

+ V (r) . (1.26)

The Hamiltonian is again the sum of the kinetic and potential energy.

Hamilton’s equation for the the generalized momentum pθ becomes

pθ = −∂H∂θ

= 0 . (1.27)

Thus pθ is independent of time implying the time independence of mr2θ. We again identifythis as the z-component of the angular momentum.

Let us finally look at Hamilton’s equation for pr. We then have

∂H

∂r= − p2

θ

mr3+∂V

∂r= −pr (1.28)

Substituting the expressions for pr and pθ, and rearranging, we obtain

m(

r − rθ2)

= −∂V∂r

. (1.29)

In order to interpret this equation, we need to find an expression for the right-hand side. Tothis end, we write

r = rer , (1.30)

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CHAPTER 1. CLASSICAL MECHANICS 11

where er is unit vector in the radial direction. This can be written as

er = cos θex + sin θey . (1.31)

Differentiating r with respect to t, we obtain

r = rer + rer

= rer + θreθ , (1.32)

where

eθ = − sin θex + cos θey , (1.33)

is a unit vector in the angular direction. Differentiating r with respect to time, we get

r = er(r − rθ2) + eθ(2rθ + θr)

= er(r − rθ2) + eθ1

r

d

dt(r2θ) . (1.34)

Using the equation of motion for pθ, i.e. that r2θ is time independent, we see that the secondterm vanishes. We then obtain

mr = −∂V∂r

er , (1.35)

which is Newton’s second law in the radial direction.——————————————————————————————————

Note that the dimension of the conjugate momentum depends on the corresponding coor-dinate. If we are using cartesian coordinates, it has the well known dimension of mass timesvelocity. In the above example, it has dimension of angular momentum. Hence the termgeneralized momentum is appropriate.

1.3 Particle in external electromagnetic fields

Given the vector potential A and the scalar potential Φ, the electric and magnetic fields canbe written as

E = −∇Φ − ∂A

∂t, (1.36)

B = ∇× A . (1.37)

We next show that the Hamiltonian

H =1

2m(p − qA)2 + qΦ , (1.38)

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CHAPTER 1. CLASSICAL MECHANICS 12

where q is the electric charge of the particles, gives the correct equation of motion. Hamilton’sequations give

x =∂H

∂px

=px − qAx

m, (1.39)

px = −∂H∂x

= q(p− qA)

m

∂A

∂x− q

∂Φ

∂x, (1.40)

and similiarly for the coordinates y and z, and the generalized momenta py and pz. The firstequation shows that the generalized momentum p is different from the kinetic momentumπ = mv. The acceleration is given by the time derivative of Eq. (1.39):

mx = px − qdAxdt

.

= px − q

(

∂Ax∂x

x+∂Ax∂y

y +∂Ax∂z

z +∂Ax∂t

)

. (1.41)

We next rewrite the time derivative of the generalized momentum as

px = q

(

x∂Ax∂x

+ y∂Ay∂x

+ z∂Az∂x

)

− q∂Φ

∂x. (1.42)

Inserting Eq. (1.42) into Eq. (1.41) and reorganizing, we obtain

mx = q

[

−∂Φ∂x

− ∂Ax∂t

]

+ q

[

y

(

∂Ay∂x

− ∂Ax∂y

)

− z

(

∂Ax∂z

− ∂Az∂x

)]

= qEx + q (yBz − zBy) . (1.43)

This is simply the x-component of the Lorentz force

F = qE + qv ×B . (1.44)

Hence the Hamiltonian Eq. (1.38) gives Newton’s second law for a charged particle in anexternal magnetic field. The replacement p → p− qA in the Hamiltonian is called minimal

substitution. We shall return to it in the chapters on field theory.

1.4 Calculus of variations

We will now discuss an alternative formulation of classical mechanics, namely a variational

formulation. Since calculus of variations is not restricted to classical physics, we discuss itmore generally. Ref. [3] provides a good introduction to variational methods.

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CHAPTER 1. CLASSICAL MECHANICS 13

Let F (y(x), y′(x)) be a function of y(x) and y′ 2, where y(a) = y1 and y(b) = y2. Considerthe integral I defined by

I[y] =∫ b

aF (y, y′) dx . (1.45)

I is a function of a function, i.e. it is a functional. Generally, a functional receives a functionas argument and returns a number. The problem of variational calculus is to find a functiony0(x), which extremizes the integral I[y], and satisfying the boundary conditions y(a) = y1

and y(b) = y2. We next write an arbitrary function y(x) satisfying the boundary conditionas y(x) = y0(x) + δy(x), where δy is the deviation. Note that δy(a) = δy(b) = 0. We canthen write

I[y0 + δy] =∫ b

aF (y, y′)dx , (1.46)

We next expand I to first order in the deviation δy. This yields

I[y0 + δy] ≈∫ b

a

[

F (y0, y′0) +

∂F

∂y

y=y0

δy(x) +∂F

∂y′

y′=y′0

δy′(x)

]

dx .

Using that δy′(x) = (δy)′ and integrating the last term in the brackets by parts, we obtain

I[y0 + δy] ≈ I[y0] +∂F

∂y′

y′=y′0

δy∣

x1

x0

+∫ b

a

[

∂F

∂y

y=y0

− d

dx

∂F

∂y′

y=y0

]

δy(x) dx (1.47)

Since the variation is zero at the end points a and b, the boundary term vanishes. Since thevariation δy is arbitrary, the integrand in δI must vanish identically, i.e.

∂F

∂y

y=y0

− d

dx

∂F

∂y′

y=y0

= 0 . (1.48)

The function y0(x) must therefore satisfy the equation (1.48).

Note: We have derived a differential equation for a function F , with a single dependentvariable y(x). It can be easily generalized to a function F = F (y1, y2, .., yN) of an arbitrarynumber of dependent variables yi(x).

Note: We have extremized the integral I, but said nothing about the nature of theextremum, i.e. whether it is a minimum, a maximum, or a saddle point. In order todetermine the nature of the extremum, we must look at the second-order variation δ2I,which is found by expanding I to second order about y0(x):

I = I0 + δI + δ2I + ... (1.49)

The sign of δ2I determines the nature of the extremum.

——————————————————————————————————

2Occasionally, F is also a function of x explicitly. The arguments clearly go through.

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CHAPTER 1. CLASSICAL MECHANICS 14

Example

Let f(x) be an arbitrary function that satiefies f(a) = y1 and f(b) = y2. The infinitesimal arc

length of the curve f(x) is given by ds =√dx2 + dy2 =

1 + (dy/dx)2 dx =√

1 + (f ′)2 dx

and the total arclength between the points (a, y1) and (b, y2) is therefore

L =∫ b

ads

=∫ b

a

1 + (f ′)2 dx . (1.50)

The problem is to find the function that extremizes the distance between the two points

(a, y1) and (b, y2). The function is F =√

1 + (f ′)2, which yields

∂F

∂f= 0 , (1.51)

∂F

∂f ′=

f ′

1 + (f ′)2. (1.52)

The equation then becomes

d

dx

f ′

1 + (f ′)2

= 0 , (1.53)

Integration then gives

f ′

1 + (f ′)2= C , (1.54)

where K1 is a constant of integration. Solving this equation with respect to f ′ gives

f ′ = ±∣

C√1 − C2

. (1.55)

Integration then yields

f(x) = K1x+K2 , (1.56)

where K1 is given in terms of C and K2 is a second integration constant. K1 and K2 aredetermined by requiring f(a) = y1 and f(b) = y2. We see that the solution to the problemis straight line.

——————————————————————————————————

In the discussion and example above, we wrote y = y(x). However, we could alsoparametrize both x and y by a parameter, for example the arc length s. One would then

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CHAPTER 1. CLASSICAL MECHANICS 15

0

t

q(t)

q(t)

q

Figure 1.1: Classical trajectory (solid curve - q0(t)) and a deviating curve (dashed curve -q(t)).

write F (y, x, y′, x′, s), where x = x(s) and y = y(s). This yields ds =√

(x′)2 + (y′)2ds and

requires the simultaneous solution of two coupled equations (see Exercise 1.5.2).

We next use the result (1.48) in classical mechanics by chosing F = L and y(x) = q(t).The boundary conditions y(a) = y1 and y(b) = y2 are replaced by q(t0) = q0 and q(t1) = q1.The classical path and a deviation is shown in Fig. 1.1 in one dimension with q(t) = x(t).

This immediately yields Lagrange’s equation 3

d

dt

(

∂L

∂q

)

=

(

∂L

∂q

)

. (1.57)

In classical mechanics the functional I is denoted by S and reads

S[x(t)] =∫ t1

t0L(q, q) dt . (1.58)

It is called the action. A classical particle is then following a path, which extremizes theaction. Since the extremum normally is minimum, the variational principle is alse called theprinciple of least action.

1.5 Problems

1.5.1 Straight line as minimal distance between points I

Show that the straight line is actually a minimum.

3If there are N generalized coordinates and momenta, there are N coupled equations.

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CHAPTER 1. CLASSICAL MECHANICS 16

1.5.2 Straight line as minimal distance between points II

Reconsider the previous problem writing

I[y(s), x(s)] =∫ s1

s0F (x(s), y(s), x′(s), y′(s))ds , (1.59)

where F =√

(x′)2 + (y′)2 and x(s0) = x0, x(s1) = x1, y(s0) = y0, and y(s1) = y1. Find

the functions x(s) and y(s) that extremize the above functional. Calculate the determinantof the Hessian and conclude that the solution is indeed a minimum in agreement with theprevious exercise.

1.5.3 Extremal curves on a sphere - Great circles

Consider the two-sphere S2 in three Euclidean dimensions. The metric on the sphere givesrise to an arclength ds, which can be written as

ds =√

dθ2 + sin2 θdφ2 , (1.60)

where θ and φ are angular variables. Find the curve that extremizes the distance betweentwo points on the sphere. You can either parametrize θ = θ(s) and φ = φ(s) or writeφ = φ(θ) (or vice versa).

1.5.4 Rigid rotator

The kinetic energy of a rigid rotator with moment of inertia I is T = 12Iω2, where ω is the

angular frequency.

a) Use spherical coordinates and show that the Lagrangian can be written as

L =1

2I(

θ2 + φ2 sin2 θ)

. (1.61)

Identify the cyclic coordinate and interpret the constant of motion.

b) Find the generalized momenta and show that the Hamiltonian can be written as

H =1

2I

(

p2θ +

p2φ

sin2 θ

)

. (1.62)

1.5.5 Mathematical pendulum

Consider a mathematical pendulum of length l and mass m in a gravitational field wherethe acceleration is g.

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CHAPTER 1. CLASSICAL MECHANICS 17

a) Choose a suitable generalized coordinate and find the Lagrangian for the system.

b) Find Lagrange’s equation.

c) Calculate the generalized momentum and the Hamiltonian.

d) Derive Hamilton’s equations for the system and show they are equivalent to Lagrange’sequation.

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Chapter 2

Lorentz transformations

In this chapter, we briefly discuss the most important aspects of Lorentz transformations inthe context of field theories. More details can be found in the chapter four in the excellenttextbook by J. B. Hartle [4].

2.1 Boosts

s

s’

v

Figure 2.1: A boost from S to S ′ along the x-axis with speed v.

Let S and S ′ be two inertial frames where S ′ moves with speed v relative to S along thex-axis, see. Fig. 2.1. Let (x, y, z, t) and (x′, y′, z′, t′) be the spacetime coordinates in the twoframes. The clocks are synchronized such that the origin of O of S coincides with the originO′ of S ′ for the t = t′ = 0. The points in spacetime are events, i. e. “something happens” attime t at a specific point in space. An example of an event is the emission of light at time tby a source located at (x, y, z).

18

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CHAPTER 2. LORENTZ TRANSFORMATIONS 19

The coordinates in the two inertial frames are related by the following transformations

x′ = γ(x− vt) , (2.1)

y′ = y , (2.2)

z′ = z , (2.3)

t′ = γ(t− vx/c2) , (2.4)

where c is the speed of light and γ = 1/√

1 − v2/c2. The transformation between S and S ′

is called a boost along the x-axis. There is of course nothing special about the x-axis andso we can define a boost in any direction given by the velocity v of S ′ relative to S. Thetransformation (2.1) and (2.4) can be conveniently written in matrix form

(

x′

ct′

)

=

(

cosh θ − sinh θ− sinh θ cosh θ

)(

xct

)

, (2.5)

where cosh θ = γ and sinh θ = γv/c. This gives

tanh θ =v

c. (2.6)

Note that both the coordinates x and t are involved in the transformation (2.1) and (2.4).This has important consequences: observers in S ′ and S do not (necessarily) agree that twoevents are taking place at the same time. The concept of simultaneous events is not absolute.More specifically, Lorentz transformations leads to

• Lorentz contraction: A rod of length L at rest is shorter by a factor 1/γ when it ismoving with speed v relative to the observer.

• Time dilation: A moving clock is ticking at a slower rate than a clock at rest.

Since space and time are intertwined, we introduce the concept of spacetime or Minkowskispace with coordinates (x, y, z, t). The points in spacetime are thus events.

The determinant of the matrix in Eq. (2.5) is unity. This is the same as for a properrotation matrix in Euclidean plane R2 which reads

(

x′

y′

)

=

(

cos θ − sin θsin θ cos θ

)(

xy

)

. (2.7)

If we write ct = iT , the matrix (2.5) is transformed into the matrix (2.7). Thus a boost canbe interpreted as a rotation. Finally, we mention that the set of all boosts defined by thevelocity v and proper rotations around the three axes in space defined by the angles θ1, θ2,and θ3 constitute the set of all homogeneous Lorentz transformations. If we add to thesetransformations, the translations in space x′ = x+c and time t′ = t+ c1, the transformationare called inhomogenoues Lorentz transformations.

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CHAPTER 2. LORENTZ TRANSFORMATIONS 20

The Newtonian limit of the transformations Eqs. (2.1)-(2.4) is found by considering veloc-ities v which are small compared to the speed of light, i.e. v ≪ c. Taking the limit c→ ∞,i.e. setting γ = 1, we obtain

x′ = (x− vt) , (2.8)

y′ = y , (2.9)

z′ = z , (2.10)

t′ = t . (2.11)

This set of transformation is known as a Galilean transformation and is the nonrelativisticlimit of special relativity. Note in partcular that the idea of simultaneity is absolute as t′ = t.The Galilean transformations are underlying Newtonian mechanics.

2.2 Geometry of Minkowski space

Let (t0, x0, y0, z0) and (t1, x1, y1, z1) be two points in S We next introduce the distance (∆s)2

between the two points in Minkowski space by

(∆s)2 = c2(∆t)2 −[

(∆x)2 + (∆y)2 + (∆z)2]

, (2.12)

where ∆t = t1 − t0 is the time difference between the events and ∆x = x1 − x0 etc. In theinertial frame S ′, we obtain

(∆s)2 = c2(∆t′)2 −[

(∆x′)2 + (∆y′)2 + (∆z′)2]

= γ2c2[

(∆t)2 − 2v∆x∆t

c2+v4

c4(∆x)2

]

− γ2[

(∆x)2 − v2(∆t)2 − 2∆x∆t]

−(∆y)2 − (∆z)2

= γ2c2(∆t)2

[

1 − v2

c2

]

− γ2(∆x)2

[

1 − v2

c2

]

− (∆y)2 − (∆z)2

= c2(∆t′)2 −[

(∆x′)2 + (∆y′)2 + (∆z′)2]

, (2.13)

where we in the last line have used that γ = 1/√

1 − v2/c2. Thus the quantity (∆s)2 isthe same in the two frames. In fact one can show that this quantity is invariant under allLorentz transformations and it can therefore be viewed as a geometric quantity independentof coordinate system 1. In differential form , we write

ds2 = c2dt2 − dx2 − dy2 − dz2 . (2.14)

We next define xµ = (ct,x), where µ = 0, 1, 2, 3. In other words x0 = ct, x1 = x, x2 = y,and x3 = z. Moreover, we introduce the metric tensor gµν which can be written as a 4 × 4

1It is basically the same concept of invariant distance between two points that you know from Euclideanspace.

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CHAPTER 2. LORENTZ TRANSFORMATIONS 21

matrix

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

. (2.15)

The shorthand notation for the metric tensor is diag(1,−, 1,−1,−1). We can then write

ds2 = gµνdxµdxν , (2.16)

where we are using the Einstein summation convention: One is summing over repeatedindices where one index is upstairs and one is downstairs. Finally note that some textbooksdefine the invariant distance as minus Eq. (2.14). The metric tensor is then defined asdiag(−1, , 1, 1, 1). This convention is just a convention and has no physical significance.

We next introduce the concept of a contravariant vector which has four components Aµ

that transform like dxµ under Lorentz transformations. Moreover, the covariant vector Aµis defined by

Aµ = gµνAν . (2.17)

For example, xµ = (ct,x) is a contravariant vector and xµ = (ct,−x) is the correspondingcovariant vector. The invariant distance ds2 can now be written as

ds2 = dxµdxµ , (2.18)

and can be considered as a scalar product between a contravariant and a covariant vectorand has the same value in all inertial frames, i.e. it is a scalar.

Note that the distance ds2 between two points can be positive, negative, or zero. Thedistance is classified according to

(∆s)2 > 0 timelike interval , (2.19)

(∆s)2 = 0 lightlike interval , (2.20)

(∆s)2 < 0 spacelike interval . (2.21)

In the case of a timelike interval, there is enough time for a signal to propagate the two pointsin space, i.e. they are in causal contact. In the case of a spacelike interval, there is not enoughtime and so the points are not causally connected. Light rays are always travelling alongpaths in spacetime where ds2 = 0 - hence the name. This is shown in Fig. 2.2, where thetwo straight lines indicate the light cone, the paths at which light is moving. The point A isinside the light cone and is in causal contact with the origin. The point B is connected tothe origin by a light signal, while the point C is outside the light cone.

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CHAPTER 2. LORENTZ TRANSFORMATIONS 22

x

ct

A

C

B

Figure 2.2: Light cone with points inside (A), on (B), and outside (C).

2.3 Differential operators

We next discuss how the differential operator ∂/∂xµ transforms under a boost along thex-axis. Using the chain rule, we can write

∂x′=

∂x

∂x′∂

∂x+

∂t

∂x′∂

∂t, (2.22)

∂y′=

∂y, (2.23)

∂z′=

∂z, (2.24)

∂t′=

∂x

∂t′∂

∂x+∂t

∂t′∂

∂t. (2.25)

In order to calculate the partial derivatives, we need the inverse of the transformations (2.1)-(2.4) which are given by

x = γ(x′ + vt′) , (2.26)

y = y′ , (2.27)

z = z′ , (2.28)

t = γ(t′ + vx′/c2) . (2.29)

This yields

∂x′= γ

[

∂x+v

c2∂

∂t

]

, (2.30)

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CHAPTER 2. LORENTZ TRANSFORMATIONS 23

∂y′=

∂y, (2.31)

∂z′=

∂z, (2.32)

∂t′= γ

[

∂t+ v

∂x

]

. (2.33)

Comparing this with the transformation rule of xµ = (ct,−x)

−x′ = γ(−x+ vt) , (2.34)

−y′ = −y , (2.35)

−z′ = −z , (2.36)

t′ = γ(t+ vx/c2) , (2.37)

we conclude that the differential operator ∂/∂xµ transforms as a covariant vector and wetherefore denote it by ∂µ.

2.4 Proper time and rate of clocks

The invariant distance between two events is

∆s2 = c2dt2 − dx2 − dy2 − dz2 . (2.38)

This can be rewritten as

ds2 =√

1 − v2/c2dt2 , (2.39)

where v is the velocity of S ′ in S. It is important to emphasize that Eq. (2.39) is valid alsowhen the velocity is not constant. Assume an observer is moving with velocity v in the frameS and carries a clock with her. The time between two events measured by this observer isdenoted by dτ and called proper time. For such an observer, Eq. (2.39) reduced to

ds2 = dτ 2 , (2.40)

and so the equation that relates laboraroty time and proper time is

dτ =√

1 − v2/c2dt . (2.41)

This shows that moving clocks run at a slower rate.

2.5 Problems

2.5.1 Interpretation of boosts

Define T by ct = iT and show that the matrix (2.5) is transformed into the matrix (2.7).

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CHAPTER 2. LORENTZ TRANSFORMATIONS 24

2.5.2 D’Alambertian under boosts

Show that ∂µ = ∂/∂xµ transforms as a contravariant vector under boosts. How does theoperator

= ∂µ∂µ (2.42)

transform?

2.5.3 Rotation matrix

Consider a rotation in the xy-plane by an angle θ. The rotation is represented by the matrix.

(

x′

y′

)

=

(

cos θ − sin θsin θ cos θ

)(

xy

)

. (2.43)

Using the chain rule, show that

∂x′=

∂x

∂x′∂

∂y+∂y

∂x′

= cos θ∂

∂x+ sin θ

∂y(2.44)

∂y′=

∂x

∂y′∂

∂y+∂y

∂y′

= − sin θ∂

∂x+ cos θ

∂y. (2.45)

Use this to show that ∇2 is invariant under proper rotations. This result can be easilygeneralized to rotations around any axis.

2.5.4 Spacelike and timelike intervals

Show that if two events are separated by a spacelike interval, we can always find an inertialframe such that ∆t′ = 0, i.e. they happen at the same time (but at different points in space).Similarly, show that of two events are separated by a timelike interval, we can always finda frame such that ∆x′ = 0, i.e. they happen at the same point in space (but at differenttimes).

2.5.5 Lorentz contraction

The length of a rod in an inertial frame is defined by two simulatenous events at its ends.The length of a rod is L∗ at rest. Assume the rod is moving with speed v along the x-axisin the inertial frame S. Show that its length in S is L = L∗/γ, i.e. it is Lorentz contracted.

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CHAPTER 2. LORENTZ TRANSFORMATIONS 25

2.5.6 Compositions of boosts

Let S, S ′, and S ′′, be three inertial frames. S and S ′ are connected by a boost along thex-axis with velocity v, and S ′ and S ′′ are connected by a boost along the x-axis with velocityv′. Show that S and S ′′ are connected by a boost along the x-axis and find the velocity v′′.

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Chapter 3

Classical field theory

In this chapter we discuss some aspects of classical field theory. A more thorough treatmentcan be found in Ref. [5].

A field φ is a mapping from spacetime M with coordinates x, y, z, and t to a field space

or a target space T :

φ : M → T . (3.1)

There are many examples of a field. Below we list a few examples:

• ρ = ρ(x, y, z, t) is the density of a liquid. ρ is a scalar function.

• The Maxwell field Aµ(x, y, z, t). The field space is that of four-vector fields.

• Quantum mechanics: ψ(x, y, z, t) is the Schrodinger wavefunction. The field space isthat of complex functions.

For each ~r, we can view φ = φ(~r, t) as a function of time t and φ therefore represents onedegree of freedom, just like q = q(t) for point particles. A field thus has infinitely manydegrees of freedom and the theory of fields replaces the theory of point particles when it isnecessary to describe the system with a continuum of degrees of freedom. Moreover, we oftenspeak of field theory in D = d+ 1 dimensions, where d is the number of spatial dimensionsand D is the dimension of space-time. In particular quantum mechanics can be viewed asfield theory in zero spatial and one time dimension, i.e. field theory in 0 + 1 dimensions.

3.1 Action and equation of motion

We first discuss the Lagrangian density L, which is a function of the fields φi, their deriva-tives and possibly explicitly of the coordinates xµ. The Lagrangian density density is thegeneralization of L = L(qi(t)qi(t), t), t to a system of with infinitely many degrees of free-dom. In most cases, it is a function of xµ only through φi and ∂µφi. Thus we can write

26

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CHAPTER 3. CLASSICAL FIELD THEORY 27

L = L(φ, ∂µφ), where we for notational simplicity have suppressed the index field index i.We will often do that in the following.

The Lagrangian L is defined by

L =∫

Ω3

d3xL(φ, ∂µφ) , (3.2)

where Ω3 is a region in space. Typically Ω3 is the whole space, i.e. Ω3 = R3. The action Sof a field theory with Lagrangian density L is defined by

S[φ] =∫ t1

t0dt L

=∫

Ω4

d4xL(φ, ∂µφ) , (3.3)

where the region in Minkowski space 1 is Ω4 = Ω3 × [t0, t1]. Typically Ω4 is whole space-time. The action S is again a functional of the field φ. In Chapter one, we were looking fora trajectory q(t) that extremizes the action. Now we are looking for a field configuration φthat extremizes the action. The field is written as

φ′ = φ+ δφ , (3.4)

where φ is the field configuration that extremizes the action and δφ is a variation whichvanishes on the boundary ∆Ω4. Inserting Eq. (3.4) into the action (3.3) we obtain

S[φ+ δφ] =∫

Ω4

d4xL(φ+ δφ, ∂µ(φ+ δφ)) . (3.5)

The action S is expanded to first order in δφ. This yields

S[φ+ δφ] =∫

Ω4

d4xL(φ, ∂µφ) +∫

Ω4

d4x

[

∂L∂φ

δφ+∂L∂∂µφ

δ(∂µφ)

]

= S[φ] +∫

Ω4

d4x

[

∂L∂φ

δφ+∂L∂∂µφ

∂µ(δφ)

]

, (3.6)

where we in the last line have used that δ(∂µφ) = ∂µ(δφ). The second term in Eq. (3.6) isdenoted by δS and reads

δS =∫

Ω4

d4x

[

∂L∂φ

δφ+∂L∂∂µφ

∂µ(δφ)

]

=∫

Ω4

d4x

[

∂L∂φ

− ∂µ∂L∂∂µφ

]

δφ+ δφ∂L∂∂µφ

δΩ4

, (3.7)

where we in the last term have integrated by parts. Since we allow no variation on thesurface ∆Ω4, the boundary term is zero. The variational principle says that δS = 0 and sowe obtain

Ω4

d4x

[

∂L∂φ

− ∂µ

(

∂L∂(∂µφ)

)]

δφ = 0 . (3.8)

1We restrict ourselves to flat spacetime, but the formalism can be generalized to curved spacetime.

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CHAPTER 3. CLASSICAL FIELD THEORY 28

Since the variation δφ is arbitrary, the term in the paranthesis must vanish, i.e. we find

∂L∂φ

= ∂µ

(

∂L∂(∂µφ)

)

. (3.9)

This is the Euler-Lagrange equation for the field φ. If the Lagrangian is a function of severalfields φi, there is an equation for each - just like we have one equation for each qi(t).

——————————————————————————————————

Example

The Lagrangian for a real scalar field is

L =1

2(∂µφ)(∂µφ) − 1

2m2φ2

=1

2gµν(∂

νφ)(∂µφ) − 1

2m2φ2 . (3.10)

This yields

∂L∂φ

= −m2φ , (3.11)

∂L∂(∂µφ)

= ∂µφ . (3.12)

The equation of motion then becomes[

∂µ∂µ +m2

]

φ = 0 . (3.13)

This can be written as[

∂2

∂t2−∇2 +m2

]

φ = 0 . (3.14)

This is the Klein-Gordon equation that describes spin-zero neutral particles of mass m as weshall see later. Note that the operator ∂µ∂

µ is called the D’Alembertian and is sometimesdenoted by .

We can also obtain the Klein-Gordon equation from a different path. Consider the energy-momentum relation of a relativistic particle with mass m:

E2 = m2 + p2 , (3.15)

In quantum mechanics, we replace the energy E and momentum p by the operators i∂0 and−i∇. Inserting this into Eq. (3.15), we obtain

− ∂2

∂t2= m2 −∇2 , (3.16)

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CHAPTER 3. CLASSICAL FIELD THEORY 29

which is equivalent to the Klein-Gordon equation (3.14).

——————————————————————————————————

The above example shows that by chosing L as in Eq. (3.10) we obtain Eq. (3.16) whichis motivated by the substitutions E → i∂0 and p → −i∇. In general, we have to writedown a Lagrangian whose equation of motion is the correct one. A general requirement to aLagrangian is that it be Lorentz invariant, i.e. it is a scalar under Lorentz transformations.Since φ is a scalar, we know that ∂µφ is a covariant vector and ∂µφ is a contravariant vector.The kinetic term is therefore a scalar since it is the scalar product between a covariant vectorand a contravariant vector. The requirement that the Lagrangian is Lorentz invariant alsoforbids a term like

φ3∇φ , (3.17)

since it is not invariant under rotations. However, it does admit a term like φ2(∇φ)2 and isan interaction term since it it fourth order in the field.

Is the Lagrangian density for a field theory unique? By that, we mean are there morethan one L that gives rise to the same equation of motion? The answer is no. We canobviously always add a constant to the Lagrangian without changing the field equations.We can also multiply the Lagrangian by a constant C since this implies that δS ′ = CδS andthat the equation of motion is not altered. Finally, we can always add the integral of a totaldivergence ∂µM

u to the action. If

S ′ = S +∫

Ω4

d4x ∂µMu . (3.18)

the second term on the right-hand side can be rewritten using Gauss’ theorem∫

Ω4

d4x ∂µMu =

∂Ω4

d3xn · M , (3.19)

where n is a normal vector to the surface Ω3 Since Mµ is not suppose to vary on the surfaceΩ3, we have δS ′ = δS and so the equation of motion is not changed.

3.2 Conjugate momentum and Hamiltonian

The conjugate momentum density πi of a field φi is defined by

πi =∂L∂φi

. (3.20)

The Hamiltonian density is then defined in analogy with the particle case

H = πiφi − L . (3.21)

——————————————————————————————————

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CHAPTER 3. CLASSICAL FIELD THEORY 30

Example

Reonsider the Lagrangian of a real scalar field. The conjugate momentum follows directlyfrom Eq (3.12) setting µ = 0:

π = φ (3.22)

The Hamiltonian density then becomes

H = πφ− L= φ2 − 1

2(∂µφ)(∂µφ) +

1

2m2φ2

=1

2

(

∂φ

∂t

)2

+1

2(∇φ)2 +

1

2m2φ2

=1

2π2 +

1

2(∇φ)2 +

1

2m2φ2 , (3.23)

where we have eliminated φ in favor of π. The Hamiltonian density is often referred to asthe energy density. Note that all terms in H are positive and so the energy density is too.

——————————————————————————————————

3.3 Symmetries and conservation laws

A coordinate qi is cyclic if the Lagrangian L does not depend explicitly on it, i.e. if

∂L

∂qi= 0 . (3.24)

Lagrange equation for qi then yields

d

dt

∂L

∂qi= 0 . (3.25)

Integrating with respect to time gives

∂L

∂qi= C , (3.26)

where C is an integration constant. In other words, a cyclic coordinate implies a conservedquantity.

We are now going to consider conserved quantities in general. More specifically we aregoing to investigate the relationship between continuous symmetries of a system and conser-vation laws in classical field theory. This is summarized in Nother’s theorem.

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CHAPTER 3. CLASSICAL FIELD THEORY 31

An example of a discrete field transformation is φ→ −φ. The Lagrangian for a real scalarfield is

L =1

2(∂µφ)(∂µφ) − 1

2m2φ2

is clearly invariant under this tranformation. An example of a continuous field transformationis Φ → Φeiα, where Φ is a complex field and α is a continuous parameter. This transformationis called a global2 phase transformation. The Larangian

L = (∂µΦ)∗ (∂µΦ) −m2Φ∗Φ , (3.27)

is invariant under global phase transformations.

Assume the Lagrangian L is invariant under a transformation of the field φ. In infinites-imal form this transformation can be written as

φ(x) → φ′(x) = φ(x) + α∆φ(x) , (3.28)

where ∆φ is a deformation of the field φ and α is an infinitesimal parameter. The Lagrangianis changing under the transformation (3.28)

L → L′ = L + α∆L , (3.29)

where

∆L =∂L∂φ

∆φ+∂L

∂ (∂µφ)∂µ (∆φ)

= ∂µ

[

∂L∂ (∂µφ)

∆φ

]

+

[

∂L

∂φ− ∂µ

(

∂L∂ (∂µφ)

)]

∆φ . (3.30)

The second term vanishes due to the equation of motion (3.9) and so we can write

∆L = ∂µ

[

∂L∂ (∂µφ)

∆φ

]

= 0 . (3.31)

This is implies a continuity equation

∂µjµ = 0 , (3.32)

where the four-current jµ is

jµ =∂L

∂ (∂µφ)∆φ . (3.33)

2It is a global phase transformation since it the same at every point in spacetime.

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CHAPTER 3. CLASSICAL FIELD THEORY 32

Integrating Eq. (3.32) over all space, we find

∂µjµ d3x =

d

dt

j0 d3x+∫

∇ · j d3x

= 0 . (3.34)

Using Gauss’ theorem, the last term can be converted into a surface integral that vanishes 3.We can then write

dQ

dt= 0 , (3.35)

where the charge Q is defined by the integral over space of the charge density ρ = j0:

Q =∫

j0 d3x . (3.36)

Eq. (3.36) is the essence of Noether’s theorem: A continuous symmetry gives rise to thecharge Q being constant in time. The conserved charge can be e.g. electric charge, baryoncharge, isospin charge or simply particle number.

——————————————————————————————————

Example

Consider the Lagrangian for a complex scalar field

L = (∂µΦ)∗ (∂µΦ) −m2Φ∗Φ . (3.37)

The Lagrangian is invariant under global phase transformations, i.e. φ → φeiα, where α isconstant. In infinitesimal form, we have

Φ → Φ + iαΦ , (3.38)

Φ∗ → Φ∗ − iαΦ∗ . (3.39)

Moreover

∂L∂ (∂µΦ)

= ∂µΦ∗ (3.40)

∂L∂ (∂µΦ∗)

= ∂µΦ . (3.41)

This yields

jµ = i [Φ (∂µΦ∗) − Φ∗ (∂µΦ)] . (3.42)

3We assume the fields and their derivatives vanish in spatial infinity.

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CHAPTER 3. CLASSICAL FIELD THEORY 33

You can use the Klein-Gordon equation for Φ directly to confirm that the jµ above satisfiesa continuity equation.

——————————————————————————————————

We can generalize the result (3.31) by noting that the Euler-Lagrange equations are notaffected by adding a total divergence to the Lagrangian 4. If the change of the Lagrangian∆L can be written as a total divergence α∂µJ µ, the transformation φ → φ + α∆φ stilldefines a symmetry. We can then write

∂µ

[

∂L∂ (∂µφ)

∆φ

]

= ∂µJ µ , (3.43)

and so

∂µjµ = 0 , (3.44)

where

jµ =

[

∂L∂ (∂µφ)

∆φ

]

−J µ , (3.45)

——————————————————————————————————

Example

Let us consider translations in space-time, which can be written as

xµ → (xµ)′

= xµ + aµ , (3.46)

where aµ is infinitesimal. To first order in aµ, we obtain

φ(x) → φ′(x)

= φ(x+ a)

= φ(x) + aµ∂µφ , (3.47)

∂µφ → ∂µφ+ aν∂µ∂νφ . (3.48)

Inserting these expressions into the Lagrangian (3.10). To first order in aµ, the Lagrangianchanges as

L → L + ∆L= L + aν∂µ(δ

µνL) . (3.49)

4In other words, changing the action by a surface term does not change the equation of motion.

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CHAPTER 3. CLASSICAL FIELD THEORY 34

The change is therefore a total divergence and this gives rise to four conserved quantities 5

The components are denoted by

T µν =

∂L∂ (∂µφ)

∂νφ− δµνL , (3.50)

where we have used that ∆φ = aµ∂νφ, cf. Eq. (3.47). T µν is called the energy-momentum

tensor. The four conserved quantities are energy and momentum and correspond to trans-lational invariance in time and space. Field theories whose Lagrangian only depend onthe coordinates xµ implicitly via φ and ∂µφ are translationally invariant just like the La-grangian (3.10).

Let us calculate the various components of T µν for the Lagrangian (3.10).. For ν = 0, we

obtain

T µ0 = (∂µφ)(∂0φ) − δµ0L . (3.51)

In particular

T 00 = (∂0φ)(∂0φ) −L

=1

2(∂0φ)(∂0φ) +

1

2(∇φ)2 +

1

2m2φ2 . (3.52)

This is the Hamiltonian density! The conservation of energy results from the invariance ofthe Lagrangian under translation in time. The spatial components are

T i0 = (∂iφ)(∂0φ) . (3.53)

The continuity equation then becomes

∂0T 00 + ∂jT j

0 = ∂0φ

[

∂2

∂t2−∇2 +m2

]

φ

= 0 , (3.54)

by virtue of the equation of motion. Furthermore, for ν = j, we obtain

T 0j = (∂0φ)(∂jφ) , (3.55)

which is the momentum density in the j-direction. Thus, momentum conservation resultsfrom the invariance of the Lagrangian under spatial translations. Finally, the spatial com-ponents are easily computed and we obtain

T ij = (∂iφ)(∂jφ) − δijL . (3.56)

——————————————————————————————————

5The four quantities aµ can varied independently.

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CHAPTER 3. CLASSICAL FIELD THEORY 35

3.4 Complex Klein-Gordon equation and gauge trans-

formations

The Lagrangian is given by

L = (∂µΦ∗) (∂µΦ) −m2Φ∗Φ , (3.57)

where Φ is a complex field. In the previuous chapter we showed that the Lagrangian for acomplex scalar field is invariant under global phase transformations i.e. transformations ofthe form

Φ → eıαΦ; , (3.58)

Φ∗ → e−ıαΦ∗ , (3.59)

where α is a constant phase, i. e. independent of spacetime coordinates. This symmetryimplies the conservation of particle number or conservation of charge.

Let us now generalize the transformations to local phase transformations, i.e. transfor-mations where α = α(x). The Lagrangian then transforms as

L → L− iΦ∗ (∂µΦ) (∂µα) + iΦ (∂µΦ∗) (∂µα) + Φ∗Φ (∂µα) (∂µα) . (3.60)

The Lagrangian is therefore not invariant under local phase transformations. However, thiscan be remidied by introducing another field Aµ, called a gauge field. The Lagrangian mustbe modified and is given by

L = (∂µ − iqAµ)Φ∗ (∂µ + iqAµ) Φ −m2Φ∗Φ , (3.61)

where q is a constant (the charge) and Dµ = ∂µ + iqAµ is called the covariant derivative.The field Aµ must transform as

Aµ → Aµ −1

q∂µα . (3.62)

The transformation (3.62) is called a gauge transformation. The Lagrangian is invariantunder the simultaneous transformation (3.62) and

Φ → eıα(x)Φ; , (3.63)

Φ∗ → e−ıα(x)Φ∗ . (3.64)

This can be seen by inserting the transformed field into the Lagrangian:

L = (∂µ − iqAµ) Φ∗ (∂µ + iqAµ)Φ −m2Φ∗Φ

→ L′

=

[

∂µ − iq

(

Aµ −1

q∂µα

)]

e−iα(x)Φ∗

[

∂µ + iq

(

Aµ − 1

q∂µα

)]

eiα(x)Φ −m2Φ∗Φ

= L . (3.65)

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CHAPTER 3. CLASSICAL FIELD THEORY 36

Electromagnetism

We next consider the Maxwell field. The electric and magnetic fields can be expressed interms of the gauge field Aµ = (φ,A) as

E = −∇Φ − ∂A

∂t, (3.66)

B = ∇× A . (3.67)

We recall that the physical fields E and B are gauge invariant, i.e. invariant under thetransformation (3.62) of 4-vector field Aµ. Maxwell’s equation are

∇ · B = 0 , ∇× E +∂B

∂t= 0 (3.68)

∇ · E = ρ , ∇×B − ∂E

∂t= j , (3.69)

The first homogenous equation simply states that we can express B as the curl of a vectorpotential A. Using this. the second homogeneous equation states that E + ∂A/∂t can bewritten as the gradient of of a scalar function Φ = A0 6. The Lagrangian we are interestedin must give rise to Maxwell inhomogeneous equations as field equations. The Lagrangiancan not simply be a found by replacing φ by Aµ in the Lagrangian for the scalar field. Doingso, yields 7

L =1

2(∂µAν) (∂µAν) − 1

2m2AµA

µ . (3.70)

The problem is that none of these terms are gauge invariant. In particular the mass term isnot gauge invariant and this explains why the photon is massless: Maxwell theory is gaugetheory and gauge invariance forbids a mass term in the Lagrangian. However, the field

strength F µν defined by

F µν = ∂µAν − ∂νAµ , (3.71)

is gauge invariant (show it!). The field strength is an antisymmetric tensor of rank two. Nowrecall that a Lagrangian must be a Lorentz scalar and therefore we must contract the twoindices upstairs. We can contract it with Fµν and try

L = −1

4FµνF

µν . (3.72)

This Lagrangian is Lorentz invariant and gauge invariant. The field strength tensor reads

F µν =

0 −Ex −Ey −EzEx 0 −Bz By

Ey Bz 0 −Bx

Ez −By Bx 0

(3.73)

6The curl of a divergence of a vector field and the gradient of the curl of a vector field always vanish.7Note that Aµ = (Φ,−A).

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CHAPTER 3. CLASSICAL FIELD THEORY 37

Maxwell’s equations can then be written as (Exercise 3.6.3)

∂µFµν = jν . (3.74)

Using the fact that Aµ = (φ,−A) it is straightforward to express Fµν in terms of E and B.It can then be shown that

F µνFµν = −2(E2 −B2) . (3.75)

In other words, the Lagrangian can be written as follows in the terms of the physical fieldsE and B.

L =1

2(E2 −B2) . (3.76)

Finally, the Lagrangian for a complex scalar field coupled to the electromagnetic field is thengiven by

L = (DµΦ∗)(DµΦ) −m2Φ∗Φ − 1

4FµνF

µν . (3.77)

3.5 Nonrelativistic limit of fields

Consider a real selfinteracting relativistic field φ whose Lagrangian is

L =1

2(∂µφ) (∂µφ) − 1

2m2φ2 − λ

24φ4 . (3.78)

In order to find the nonrelativistic limit, we write the field as

φ =1√2m

(

eimtψ + e−imtψ†)

. (3.79)

Inserting the representation Eq. (3.79) into Eq. (3.78) and omit the oscilating terms (i.e. theterms that go like e±2imt), we obtain

1

2m2φ2 =

1

4m(

2ψ†ψ + e2imtψψ + e−2imtψ†ψ†)

≈ 1

2mψ†ψ , (3.80)

where ≈ implies that we have omitted the oscillating terms. Similarly, we obtain

1

2(∂0φ)(∂0φ) ≈ 1

2mψ†ψ† +

i

2

(

ψ†∂0ψ − ψ∂0ψ†)

(3.81)

1

2(∂iφ)(∂iφ) ≈ − 1

2m(∇ψ†) · (∇ψ) (3.82)

λ

24φ4 ≈ λ

(4m)2(ψ†ψ)2 , (3.83)

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CHAPTER 3. CLASSICAL FIELD THEORY 38

Putting the pieces together, we obtain

Lnr =i

2

(

ψ†∂0ψ − ψ∂0ψ†)

− 1

2m(∇ψ†) · (∇ψ) − 1

2g(ψ†ψ)2 , (3.84)

where we have defined g = λ/8m2. Integrating the first term by part, we obtain the standardform of a nonrelativistic field theory:

Lnr = iψ†∂0ψ − 1

2m(∇ψ†) · (∇ψ) − 1

2g(ψ†ψ)2 . (3.85)

The Lagrangian density Eq. (3.85) is used as a starting point for nonrelativistic selfinteractingfield theories.

3.6 Problems

3.6.1 Nonrelativistic field theory

The Lagrangian density for the nonrelativistic field ψ is given by

L = −1

2ih(ψ∗ψ − ψ∗ψ) − h2

2m(∇ψ∗)·(∇ψ) . (3.86)

1) Find the equation of motion for ψ.2) Find the Hamiltonian density H.3) Show that the Lagrangian density is invariant under a global phase transformation.4) Find the conserved current that corresponds to the symmetry and interpret the continuityequation.5) How can you make the Lagrangian density invariant under local phase transformations?

3.6.2 Dual field tensor

The dual tensor F µν is defined by

F µν =1

2εµναβFαβ , (3.87)

where εµναβ is the Levi-Civita symbol. εµναβ is totally antisymmetric under permutation ofindiced and ε0123 = +1. Find the matrix F µν and show that

F µνFµν = −4E ·B . (3.88)

Finally, find an expression for F µνFµν .

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CHAPTER 3. CLASSICAL FIELD THEORY 39

3.6.3 Maxwell’s equations

The Lagrangian for the electromagnetic field with sources is given by

L = −1

4FµνF

µν − jµAµ , (3.89)

where jµ = (ρ,−j) is the external four-current. Is L gauge invariant? Find Euler-Lagrange’sequations and express them in terms of the electromagnetic field, i.e. in terms of E and B.Comments?

3.6.4 Free field theory for fermions

The Lagrangian for the free electron-positron field is

L = ψ(iγµ∂µ −m)ψ . (3.90)

1) Calculate the components of the energy-momentum tensor. Find the energy and momen-tum densities and show explicitly current conservation.

2) The Lagrangian is invariant under global phase transformations:

ψ → eiαψ (3.91)

ψ → e−iαψ (3.92)

This gives rise to the well-known conservation of charge. Consider instead the transformation

ψ → eiγ5αψ (3.93)

ψ → ψei(γ5)†α , (3.94)

where γ5 = iγ0γ1γ2γ3 is a matrix. The transformation is called an chiral transformation.Show that

(γ5)† = γ5 , (3.95)

(γ5)2 = 1 , (3.96)

γ5, γµ = 0 . (3.97)

3) For what values of m is the chiral transformation a symmetry of the Dirac Lagrangian?Find the corresponding 4-current (called an axial vector current since it is changes sign underparity, x → −x).

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Chapter 4

Relativistic Wave equations

4.1 Klein-Gordon equation

See Hemmer chapter 17.

4.1.1 Relativistic corrections

In this chapter, we briefly discuss the leading relativistic corrections to the energy of a scalarparticle in a static Coulomb field.

The energy E of a particle with mass m and momentum p is given by

E =√

m2c4 + p2c2

= mc2√

1 +(

p

mc

)2

. (4.1)

For small momenta, i.e. for momenta p ≪ mc, we can expand Eq. (4.1) in a power seriesaround p = 0. To second order, we obtain

E ≈ mc2[

1 +1

2

(

p

mc

)4

− 1

8

(

p

mc

)4

+ ...

]

= mc2 +p2

2m− p4

8m3c2. (4.2)

The first term is the energy associated with the rest mass of the particle. The second termis the usual kinetic term and is included in the standard nonrelativistc Hamiltonian H0. Fora particle in a static Coulomb field, this is

H0 = − h2

2m∇2 − Ze2

4πǫ0r. (4.3)

40

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 41

The third term is the first relativistic correction to the energy. We will denote this term byH ′. In the coordinate representation, we have p = −ih∇ and is

H ′ = − h4

8m3c2∇4 . (4.4)

We know that T = p2

2m= − h2

2m∇2 commutes with the total angular momentum L2 as well as

the x-component Lz, i.e.

[T, L2] = [T, Lz ] = 0 . (4.5)

Since H ′ ∼ T 2, this implies

[H ′, L2] = [H ′, Lz] = 0 . (4.6)

Furthermore, H ′ does not act in spin space and so

[H ′, S2] = [H ′, Sz] = 0 . (4.7)

This in turn implies that H ′ is diagonal in the quantum numbers l,ml, s, and ms. Since theenergy depends only on the principal quantum number n, we must in general use degen-erate perturbation theory but due to vanishing commutators, this is not necessary for theperturbation H ′. The first-order energy shift E1 of the state |ψ〉 is then given

E1 = 〈ψ|H ′|ψ〉

= − 1

2mc2〈ψ|T 2|ψ〉 . (4.8)

We can now write

T = H0 +Ze2

4πǫ0r, (4.9)

where H0 is the unperturbed Hamiltonian. This yields

E1 = − 1

2mc2〈ψ|

(

H0 +Ze2

4πǫ0r

)2

|ψ〉

= − 1

2mc2

E20 + 2E0

Ze2

4πǫ0

1

r

+

(

Ze2

4πǫ0

)2 ⟨1

r2

, (4.10)

where is the unperturbed energy:

E0 = −mc2Z2α2

2

1

n2. (4.11)

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 42

In order to proceed, we need to know the expectation values of 1/r and 1/r2 in the unper-turbed state. These are given by1

1

r

=mcZα

2h

1

n2, (4.12)

1

r2

=m2c2Z2α2

h2

1

n3(l + 12). (4.13)

We can then write the energy to first order in perturbation theory

E = mc2 + E0 + E1

= mc2 − mc2Z2α2

2

1

n2+mc2Z4α4

n4

(

3

8− n

2l + 1

)

, (4.14)

where we have used α = e2/(4πǫ)hc. Note that the shifted energy depends on the angularquantum number l and so the degeneracy in l from Schrodinger theory is lifted once oneincludes relativistic corrections. Eq. (4.14) is obtained by expanding the relativistic energy

E =mc2

1 + Z2α2

(n−l−1/2+√

[l+1/2)2−Z2α2)]2

(4.15)

to second order in Z2α2.

4.2 Dirac equation

In this section, we discuss some extra aspects of the Dirac equation not covered in thetextbook.

4.2.1 Representation independence

Given two sets of γ-matrices which both satisfy

γµγν = 2gµν (4.16)

Then there exists a nonsingular matrix S such that 2

(γ′)µ = SγµS−1 . (4.17)

Next assume that ψ is a solution to the Dirac equation, i.e.

(iγµ∂µ −m)ψ = 0 . (4.18)

1See e.g. Hemmer Sec. 5.7.2Have confidence in this statement!

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 43

Inverting Eq. (4.17), we obtain γµ = S−1(γ′)µS. This yields

(iS−1(γ′)µS −m)ψ = 0 . (4.19)

Multiplying this equation with S from the left, we obtain

(i(γ′)µ −m)Sψ = 0 . (4.20)

In other words, ψ′ = Sψ is a solution to the Dirac equation with the new γ-matrices. Thisshows the representation independence. Note: The components in the two wavefunctions ψand ψ′ are not the same since the matrix S shuffle them around, ψ′ = Sψ.

4.2.2 Relativistic corrections

We now consider the Dirac equation in an external electromagnetic field given by Aµ and itsnonrelativistic limit. A complete discusion can be found in Refs. [8, 17]. The Dirac equationfor stationary states in an external field Aµ reads

i (γµ∂µ + iqAµ)ψ −mψ = 0 , (4.21)

where we have replaced the partial derivative by the covariant derivative. This can berewritten as

γ0 (i∂0 − qA0)ψ = γj (−i∂j + qAj)ψ +mψ . (4.22)

Assuming the time-dependence e−iEt for the stationary state ψ and using p = −i∇, we canwrite

γ0 (E − qA0)ψ = γ · (p − qA)ψ +mψ . (4.23)

If we introduce two two-component vectors ψA and ψB and using the standard representationof the γ-matrices, we can write the Dirac equatiton in an external field as follows

(

E − qA0 −m −σ · (p− qA)σ · (p− qA) −(E − qA0 +m)

)(

ψAψB

)

= 0 . (4.24)

The fields ψA and the field ψB are referred to as the large and small components of the fieldψ, respectivly. The reason is that the components in ψA are much larger than those of ψBin the nonrelativistic limit. Eq. (4.24) can be written as

−σ · (p− qA)ψA = (E − qA0 +m)ψB (4.25)

−σ · (p − qA)ψB = (E − qA0 −m)ψA . (4.26)

We can use the second of these equations to eliminate the ψB from the first and the equationfor the large component ψA can now be written as

[σ · (p− qA)]

[

1

E − qA0 +m)

]

[σ · (p − qA)]ψA = (E − qA0 −m)ψA . (4.27)

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 44

We next write E = m + ENR and assume E − m = ENR ≪ m and |qA0| ≪ m. This issimply the nonrelativistic approximation. We obtain

1

2m[σ · (p− qA)]2ψA = (ENR − qA0)ψA (4.28)

ψB = −σ · (p − qA)

E +m− qA0ψA . (4.29)

Reinstating factors of c, one can show from Eq. (4.29) that ψB ≈ v/cψA. Using the identity

(σ · A)(σ · B) = A · B + iσ(A ×B) (4.30)

the operator in Eq. (4.28) we obtain

[σ · (p − qA)]2 = (p− qA)2 + iσ · [(p − qA)] × (p− qA)]]

= (p− qA)2 − qiσ · [p× A + A × p]

= (p− qA)2 − qiσ · B , (4.31)

where we have used

p× A + A × p = −i(∇×A)

= −iB . (4.32)

This yields the nonrelativistic Hamiltonian

HNR =1

2m(p− qA)2 + qA0 −

q

2mσ · B , (4.33)

and the Schrodinger equation

HNRψA = ENRψA . (4.34)

The first term is obtained from the free part of the nonrelativistic Hamiltonian for a particleby minimal substitution, p → p−qA. For a an electron bound to a nucleus, the second termis the usual electrostatic energy of a point charge. The last term is the interaction of thespin of the electron with the external magnetic field. A magnetic moment µ in an externalmagnetic field is E = −µ · B. If we write µ = gsq/2mσ in analogy with µ = q/2mL, weconclude that the magnetic moment of an electron is gs = 2.

We can systematically find corrections to the nonrelativistic Hamiltonian in Eq. (4.33).This is basically an expansion in v/c. We will consider three such terms. This is The firstterm is

Hc1 =1

8m3(−i∇)4 . (4.35)

This correction is arising from expanding the energy-momentum relation:

E =√

m2c4 + c2p2

= mc2 +p2

2m− p4

8m3c2+ ... , (4.36)

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 45

to fourth order in p. These terms are also present in the Klein-Gordon theory. This term isdiagonal in l, ml, and ms and so we can use first-order perturbation theory directly.

The second term is

Hc2 = −qhσ · (E × p)

4m2c2. (4.37)

For a time-independent potential A0, we have

E = −∇A0

= −dA0

drer . (4.38)

Inserting this into (4.37), we obtain

Hhc2 =qh

4m2c21

r

dA0

drσ · (r× p)

=qh

4m2c21

r

dA0

drσ · L . (4.39)

This is the well-known spin-orbit coupling. This term is not diagonal in l, ml, and ms andso we must change basis.

Finally, the third term we consider is

Hc3 = − qh2

8m2c2∇ · E . (4.40)

The right-hand side is proportional to the charge density. For a Coulomb potential we haveρ = q′δ(r), where q′ is the charge. and in this case we obtain

Hc3 = − qq′h2

8m2c2δ(r) . (4.41)

This term was first found by Charles G. Darwin (1887-1962) and is called the Darwin term.He was the grandson of the Charles Darwin (1809-1882). Note that for an electron in the fieldof a nucleus of charge q′ = Z|e|, the prefactor is positive. Since it involves the delta-functionin the origin, it affects s-states only (only s-states have nonvanishing wavefunctions in theorigin). We can use nondegenerate perturbation theory since it is diagonal in l, ml, and ms.

Exact spectrum

The energy spectrum for a Dirac fermion in a Coulomb potential can be calculated ex-actly [8]. The orbital and spin angular momentum operators do not commute with theDirac Hamiltonian and are thus not good quantum numbers. However, the sum J = L + S

does commute with H and so the quantum number j can be used to label the solutions Thespectrum is

Eexact =m

1 + Z2α2

n−j−1/2+√

(j+1/2)2−Z2α2

, (4.42)

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 46

where n is the usual principal quantum number and j is the total angular momentumquantum number. Upon expansion to second order in Z2α2, we obtain

E = m

[

1 − 1

2

(zα)2

n2− 1

2

(zα)4

n3

(

1

j + 12

− 3

4n

)]

. (4.43)

This is identical to the sum of the first-order energy shifts calculated using degenerate per-turbation theory with the three perturbations considered above.

4.3 Problems

4.3.1 Operator relations

Show the operator relation

(σ · A) (σ · B) = A · B + iσ · (A × B) . (4.44)

4.3.2 Klein-Gordon equation in external fields

The Lagrangian density of a complex scalar field coupled to the Maxwell field is

L = (DµΦ)∗(DµΦ) −m2Φ∗Φ − 1

4FµνF

µν .

1) Show that the equation of motion for Φ∗ reads

[

DµDµ +m2

]

Φ = 0 .

We next consider a particle with charge q in two spatial dimensions in an external field wherethe vector potential is given by Aµ = (0,−By, 0, 0), where B is a constant.

2) Compute the electric field E and the magnetic field B.

3) Show that the Hamiltonian density can be written as

H = (∂0Φ)∗ (∂0Φ) + [(∂x − iqBy)Φ∗] [(∂x + iqBy)Φ] + (∂yΦ)∗ (∂yΦ)

+m2Φ∗Φ +1

2B2 .

Is H Lorentz invariant?

4) The eigenfunctions can be written as

Φ = e−i(Et−pxx)f(y) ,

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CHAPTER 4. RELATIVISTIC WAVE EQUATIONS 47

where E is the energy of the state and px is the x-component of the momentum. Show byinserting Φ into the equation of motion that f(y) satifies

[

− d2

dy2− E2 +m2 + (px − qBy)2

]

f(y) = 0 .

Use this result to derive the spectrum, i.e. the energy eigenvalues. Hint: Harmonic oscilla-tor.

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Chapter 5

Field quantization

In this chapter, we discuss quantization of field. As a prelude we briefly review the harmonicoscillator and its quantization.

5.1 Harmonic oscillator

The Hamiltonian H of a one-dimensional harmonic oscillator with frequency ω and mass mis given by

H =p2

2m+

1

2mω2x2 , (5.1)

where x is the position operator of the oscillator and p is its linear momentum operator.

From h, the mass and the frequency , we can form quantity l =√

h/mω which sets thelength scale of the oscillator.

Let us define the dimensionless operators a and a† by

a =1√2

(

x

l+ip

hl)

, (5.2)

a† =1√2

(

x

l− ip

hl)

, (5.3)

It is then easy to verify that aand a† satisfy the commutation relation

[a, a†] = 1 . (5.4)

Inverting the relations (5.2) and (5.3), we obtain

x =l√2(a + a†) , (5.5)

p =h√2il

(a− a†) . (5.6)

48

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CHAPTER 5. FIELD QUANTIZATION 49

We can now express the Hamiltonian Eq. (5.1) in terms of the operators a and a†. Thisyields

H = − 1

4m

h2

l2(a− a†)2 +

1

4mω2l2(a+ a†)2

=1

2hω

(

aa† + a†a)

= hω(

1

2+ a†a

)

, (5.7)

where we in the last line have used the commutator (5.4). Note that the creation operatora† stands to the left of the annihilation operator a. Whenever we have an operator A whichis a product of a and a† and all the creation operators stand to the left of the annihilationoperators, the operator A is called normal ordered.

If we introduce the operator N = a†a, the problem of finding the eigenvalues for H is thesame as finding the eigenvalues of N , since

H = hω(

1

2+N

)

. (5.8)

——————————————————————————————————

Example

The commutator between N and a is

[N, a] = [a†a, a] = a†[a, a] + [a†a]a = −a . (5.9)

This implies that

[H, a] = −hωa . (5.10)

Similarly, one can show that

[N, a†] = a† . (5.11)

——————————————————————————————————

Let |ψE〉 be an eigenstate of the Hamiltonian H with energy E, i.e.

H|ψE〉 = E|ψE〉 . (5.12)

If a|ψE〉 also is a vector in Hilbert space, we must have

〈ψE|a†a|ψE〉 > 0 . (5.13)

Moreover

Ha|ψE〉 = a(H − hω)|ψE〉 ,

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CHAPTER 5. FIELD QUANTIZATION 50

where we have used the the commutator between a and H , Eq. (5.10). This yields

Ha|ψE〉 = a(E − hω)|ψE〉 . (5.14)

Thus a|ψE〉 is an eigenstate of H with energy E − hω. The operator a is therefore called alowering operator. This implies that

〈ψE|a†a|ψE〉 = 〈ψE|H − 1

2hω|ψE〉

= E − 1

2hω

> 0 . (5.15)

Thus E > 12hω. We can continue these arguments ad infinitum lowering the energy by

applying a. Thus the energy of the system is unbounded from below and this makes nosense. The only way out is that there exists a state, |0〉, such that a annihilates it:

a|0〉 = 0 . (5.16)

We call this state the vacuum state. Moreover, it follows from Eqs. (5.7) and (5.16)

H|0〉 = hω(

1

2+ a†a

)

|0〉

=1

2hω|0〉 . (5.17)

Finally, consider the action of H on the state a†|0〉. One finds

Ha†|0〉 = (hωa† + a†H)|0〉

=3

2hωa†|0〉 , (5.18)

where we have used [H, a†] = hωa† and Eq. (5.17). Thus the the state a†|0〉 is an eigenstateof H with eigenvalue 3

2hω. More generally, one can show that if |ψE〉 is an eigenstate of

H with energy E, then a†|ψE〉 is an eigenstate of H with energy E + hω. The operator a†

is therefore called a raising operator. The whole Hilbert space can then be built from theground state |0〉 and the raising operator a†. An excited state can be written as

|n〉 =(

a†)n |0〉 , (5.19)

whose energy is En = hω(n+ 12) and contains n quanta.

5.2 Fields

When we are quantizing a system of N particles, the commutation relations are given by

[qi, pj] = iδij , [qi, qj ] = [pi, pj ] = 0 , (5.20)

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CHAPTER 5. FIELD QUANTIZATION 51

where i, j = 1, 2, 3...N . In a field theory, we have infinitely many degrees of freedom, which isreflected in the fact that we have introduced the momentum density π, which is the conjugatevariable to the field φ. For a scalar field φ this is

π(x, t) =∂L

∂(∂0φ). (5.21)

The field φ(x, t) and its conjugate momentum π(y, t) play the role of qi and pj where x andy are continuous indices corresponding to i and j. Quantization of a field theory is nowcarried out by postulating the following commutation relations:

[φ(x, t), π(y, t)] = iδ3(x − y) , [φ(x, t), φ(y, t)] = [π(x, t), π(y, t)] = 0 . (5.22)

Two remarks are in order. Firstly, the Kronecker delta is replaced by Dirac’s delta function,which is natural given the continuous nature of a field. Secondly, note that the commutatorsare at the same time t for φ(x, t) and π(y, t), whence they are called equal time commutation

relations.

The Klein-Gordon equation

(

∂2

∂t2−∇2 +m2

)

φ(x, t) = 0 , (5.23)

admits plane-wave solutions of the form

φ ∼ e−ipx , (5.24)

where px = p0t − p · x and p0 =√

|p|2 +m2. These solutions form a complete set and wetherefore expand the classical field in a Fourier series with Fourier coefficients ap and a∗p

φ(x, t) =∫

d3p

(2π)3

1√

2Ep

[

ape−ipx + a∗pe

ipx]

, (5.25)

where Ep =√

|p|2 +m2. The prefactor 1√2Ep

is simply a convenient normalization of the

solutions. The coefficients ap and a∗p are classically Fourier coefficients, but since φ(x, t) isa quantum field they are really operators and we therefore write the field operator as

φ(x, t) =∫

d3p

(2π)3

1√

2Ep

[

ape−ipx + a†pe

ipx]

, (5.26)

The momentum density can now be writen

π(x, t) =∫

d3p

(2π)3(−i)

Ep2

[

ape−ipx − a†pe

ipx]

. (5.27)

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CHAPTER 5. FIELD QUANTIZATION 52

The equal-time commutator can now be written as

[φ(x, t), π(y, t)] =∫

d3p

(2π)3

d3q

(2π)3

(−i2

)

EqEp

[

[ap, aq]e−i(px+qy)

−[a†p, a†q]e

i(px+qy) + [a†p, aq]ei(px−qy) − [ap, a

†q]e

−i(px−qy)]

= iδ3(x − y) . (5.28)

Next we assume that the operators above satify the commutation relations[

ap, a†q

]

= (2π)3δ3(p− q) (5.29)

[ap, aq] = 0 , (5.30)[

a†p, a†q

]

= 0 . (5.31)

This yields

[φ(x, t), π(y, t)] = i(2π)3∫

d3p

(2π)3

d3q

(2π)3

EqEp

[

δ3(p− q)]

e−i(px−qy) . (5.32)

Integrating over q gives

[φ(x, t), π(y, t)] = i∫ d3p

(2π)3eip·(x−y) . (5.33)

The integral on the right-hand-side is the integral representation of the δ-function in threedimensions, i.e. δ3(x − y). Thus we recover the first equal-time commutation relation inEq. (5.22). The remaining commutators can be verified in the same manner. Now thecommutators above are simply those of the harmonic oscillator. We can therefore interpretour quantum field as a collection of independent harmonic oscillators with frequency ω = EpLet us finally note that we could have started by expanding the scalar field in Fourier seriesof eipx with Fourier coefficients ap and a∗p as in Eq. (5.25). Quantization is then carriedout directly by promoting the Fourier coefficients to operators ap and a†p satisfying thecommutation relations for the harmonic oscillator. Working out the commutator of φ(x, t)and π(x, t) we recover Eq. (5.22).

We next consider the Hamiltonian, which can be written as

H =∫

d3xH , (5.34)

where H is the Hamiltonian density

H =1

2

(

∂φ

∂t

)2

+1

2(∇φ)2 +

1

2m2φ2 . (5.35)

For a single oscillator with frequency ω, we know the Hamiltonian:

H = ω(

1

2+ a†a

)

. (5.36)

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CHAPTER 5. FIELD QUANTIZATION 53

Inserting the expression for φ and its derivatives yields

H =∫ d3p

(2π)3Ep

(

1

2[ap, a

†p] + a†pap

)

=1

2V∫ d3p

(2π)3Ep +

∫ d3p

(2π)3Epa

†pap , (5.37)

where we have used that δ3(0) = V/(2π)3 and V is the volume of space. Note that the firstterm is actually divergent since it is the integral of the ground-state energies of the harmonicoscillators with energy Ep. This suggests that the ground state energy (the vacuum energy)of a quantum field is infinite! However, we can only measure energy differences betweenstates and so it is customary to simply delete this term 1. Note, however, this may bedangerous if we are dealing with general relativity where all energy density is a source ofcurvature in Einstein’s field equations.

It is now easy to calculate the commutator between H and aq:

[H, aq] =∫ d3p

(2π)3Ep[a

†pap, aq]

=∫

d3p

(2π)3Ep[a

†p, aq]ap

= −∫

d3p

(2π)3Ep(2π)3δ3(p− q)ap

= −Eqaq , (5.38)

i.e. Haq = aq(H − Eq). This implies that Hnaq = aq(H −Eq)n and therefore

eiHtaqe−iHt = aqe

−iEqt . (5.39)

By taking the adjoint, we obtain 2

eiHta†qe−iHt = a†qe

iEqt . (5.40)

We have now found the time dependence of the creation and annihilation operators and soin the time-independent Schrodinger picture, we find

φ(x) =∫

d3p

(2π)3

1√

2Ep

[

apeip·x + a†pe

−ip·x]

, (5.41)

π(x) =∫

d3p

(2π)3(−i)

Ep2

[

apeip·x − a†pe

−ip·x]

. (5.42)

1It is related to the operator-ordering problem in quantum field theory. The expression xp is the sameas px in classical mechanics since they are functions. They are not identical once x and p are promoted tooperators.

2Recall H = H†.

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CHAPTER 5. FIELD QUANTIZATION 54

By letting p → −p in the second terms inside the bracket, these equations can be conve-niently written as

φ(x) =∫

d3p

(2π)3

1√

2Ep

[

ap + a†−p

]

eip·x , (5.43)

π(x) =∫ d3p

(2π)3(−i)

Ep2

[

apeip·x − a†−p

]

eip·x . (5.44)

The vacuum state is denoted by |0〉 and is by definition annihilated by ap, i.e.

ap|0〉 = 0 . (5.45)

A single-particle state that contains a particle with momentum p and energy Ep is given by

|p〉 = a†p|0〉 . (5.46)

If the vacuum state is normalized, i.e. if 〈0|0〉 = 1, the norm of the state |p〉 is

〈p|q〉 = 〈0|apa†q|0〉

= 〈0|a†qap + (2π)3δ3(p − q)|0〉= (2π)3δ3(p− q) . (5.47)

It can be shown that this normalization is not Lorents invariant. The problem is that volumesare not invariant. In order to make the normalization of the states invariant, one must definethe single-particle state as

|p〉 =√

2Epa†p|0〉 . (5.48)

5.3 Nonrelativistic field theories

In the preceding chapter, we have quantized the Klein-Gordon equation which is a relativisticfield theory. In this chapter, we take a closer look at nonrelativistic (NR) field theories. TheLagrangian of a free nonrelativistic theory is

L = iψ∗∂0ψ − 1

2m(∇ψ)∗(∇ψ) . (5.49)

The canonical momentum is then

∂L∂(∂0ψ)

= iψ∗

≡ Π . (5.50)

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CHAPTER 5. FIELD QUANTIZATION 55

Note that in the NR case, the conjugate momentum is essentially equal to the complexconjugate of the field and not its time derivative. This is of course due to the fact that L islinear and not quadratic in the time derivative. Moreover, we have

∂L∂(∂iψ)

=1

2m∂iψ∗ . (5.51)

The equation of motion then becomes

∂µ∂L

∂(∂µψ)= 0 , (5.52)

since ∂L∂ψ

= 0. Inserting Eqs. (5.50) and (5.51) into Eq. (5.52), we obtain the Schrodingerequation

i∂0ψ∗ − ∇2

2mψ∗ = 0 . (5.53)

The Schrodinger equation admits solution of the form

ψ(x, t) = e−i(p0t−p·x) , (5.54)

where p0 = p2/2m = Ep3. The field operator ψ(x, t) can now be expanded in a complete

set of solutions as usual 4

ψ(x, t) =∫

d3p

(2π)3ape

−i(p0t−p·x) , (5.55)

and the adjoint is given by

ψ†(x, t) =∫

d3p

(2π)3a†pe

i(p0t−p·x) , (5.56)

Again we assume the equal-time commutation relations

[ψ(x, t),Π(y, t)] = iδ3(x − y) , (5.57)

while all other commutators vanish. This yields

[ψ(x, t), ψ†(y, t)] = δ3(x − y) . (5.58)

Inserting the expressions for ψ and ψ† into the commutator, we find

[ψ(x, t), ψ†(y, t)] =∫

d3p

(2π)3

d3q

(2π)3

[

ap, a†q

]

e−i(p0−q0)tei(p·x−q·y) . (5.59)

3This relation is found by inserting the ansatz (5.54) into the Schrodinger equation.4Note that we have already promoted the Fourier coefficients to operators

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CHAPTER 5. FIELD QUANTIZATION 56

Assuming that[

ap, a†q

]

= (2π)3δ3(p − q) , (5.60)

the commutator reduces to

[ψ(x, t), ψ†(y, t)] =∫

d3p

(2π)3

d3q

(2π)3e−i(p0−q0)tei(p·x−q·y)(2π)3δ3(p − q)

=∫

d3p

(2π)3eip·(x−·y)

= δ3(x − y) , (5.61)

as it should. The other commutators are worked out in the same manner.

We have seen that the Hamiltonian can be written as

H =∫ d3p

(2π)3

1

2m∇ψ† · ∇ψ

=∫

d3p

(2π)3

p2

2ma†pap . (5.62)

Let |0〉 be the vacuum state, i.e. the state that is annihilated by aq for all q, i.e. aq|0〉 = 0.What is the state a†q|0〉? Letting H act on it, we find

Ha†q|0〉 =∫

d3p

(2π)3

p2

2ma†papa

†q|0〉

=∫

d3p

(2π)3

p2

2ma†p[

(2π)3δ3(p− q) + a†qap

]

|0〉

=q2

2ma†q|0〉 , (5.63)

where we have used that ap annihilates the vacuum and that the delta-function picks outp = q from the integral. Thus the state a†q|0〉 is an eigenstate of H with eigenvalue q2/2m.Thus it is a single-particle state with energy E = q2/2m. Not exactly shocking.

5.4 Note on fermions

If we are quantizing fermions, we cannot postulate the equal-time commutator relation (5.57).Instead we postulate the equal-time anticommutator 5

ψ(x, t),Π(y, t) = iδ3(x − y) . (5.64)

5In NR field theory it is not obvious that we need the anticommutator in the first place. However,quantizing Dirac theory using the commutator leads to logical inconsistences.

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CHAPTER 5. FIELD QUANTIZATION 57

Working out the details, one can show that the fundamental relation (5.64) is consistentwith the anticommutator between the annihilation and creation operators

ap, a†q

= (2π)3δ(p− q) , (5.65)

while all other anticommutators vanish. The anticommutator relations have some importantconsequences. If we denote, as usual, |0〉 as vacuum state, the state

a†q|0〉 , (5.66)

is a state containing a particle with momentum q. What about the state

|ψ〉 = a†pa†q|0〉 . (5.67)

Using the anticommutator, we can write

|ψ〉 = −a†qa†p|0〉.

The minus sign shows that the state |ψ〉 is antisymmetric under the permutation of the twoparticles and thus the anticommutor implements the Pauli principle. Moreover, if p = q, wefind

|ψ〉 = −|ψ〉 . (5.68)

In other words, |ψ〉 = 0. Thus a state cannot contain more than a single particle with thesame momentum. This is again the Pauli principle stating that the states in Hilbert statecannot accommodate two or more identical particles in the same single-particle state. Thusthe state |q〉 = a†q|0〉 is annihilated by a†q just like |0〉 is annihilated by aq and

ap|q〉 = apa†q|0〉

=[

ap, a†q

− a†qap

]

|0〉 (5.69)

= (2π)3δ3(p− q)|0〉 .

This implies

〈0|ap|q〉 = 〈0|apa†q|0〉

= (2π)3δ3(p− q) , (5.70)

where we have used that the vacuum state is normalized to unity.

5.5 Propagator

We next consider the following quantity

D(x− y) = 〈0|φ(x)φ(y)|0〉 . (5.71)

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CHAPTER 5. FIELD QUANTIZATION 58

The fact that the function D is a function of the difference x − y follows from Lorentzinvariance. Inserting the expressions for φ(x) and φ(y) into this equation, we obtain

D(x− y) =∫ d3p

(2π)3

d3q

(2π)3

1

2√

EpEqeiqy−ipx〈p|q〉 , (5.72)

where we haved used that ap annihilates the vacuum. Since the 〈p|q〉 = (2π)3δ3(p − q), itreduces to

D(x− y) =∫

d3p

(2π)3

1

2Epeip(y−x) . (5.73)

We next define the Feynman propagaor by

DF (x− y) =

D(x− y) , x0 > y0

D(y − x) , y0 > x0.(5.74)

Using the step function this can be rewritten as

DF (x− y) = θ(x0 − y0)D(x− y) + θ(y0 − x0)D(y − x) ,

≡ 〈0|T [φ(x)φ(y)]|0〉 , (5.75)

where T is called a time-ordered product. Note that the earliest time stands to the right. Wewill next show that the Feynman propagator can be written as a four-dimensional integral

DF (x− y) =∫

d4p

(2π)4

i

p2 −m2 + iǫe−ip(x−y) , (5.76)

where ǫ > 0 is a small quantity. The poles in the complex p0 plane are given by p0 =±(Ep− iǫ). The term iǫ ensures that the poles are slightly off the real axis. In order to showthat the above expression is correct, we must perform the integration over p0. The integralwe must evaluate is 6

I =∫

dp0

i

(p0 + Ep − iǫ)(p0 − Ep + iǫ)e−ip0(x

0−y0) (5.77)

When x0 < y0, we can close the contour in the upper half plane, see Fig. 5.1. We then pickup a contribution from the pole p0 = −Ep + iǫ. The integral then reduces to

I =1

2EpeiEp(x0−y0) . (5.78)

We next want to show that the contribution to the integral from the semi-circle vanishesin the limit where its radius R → ∞. To this end, we write the complex variable p0 as

p0 = Reiθ

= R cos θ + iR sin θ . (5.79)

6We are simply considering the p0-dependent part of the integral in Eq. (5.76).

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CHAPTER 5. FIELD QUANTIZATION 59

Figure 5.1: Contour in the complex p0-plane.

This implies that we can write

Re[eiEp(x0−y0)] = eR sin θ(x0−y0) . (5.80)

In the upper plane, we have Imp0 > 0 and since x0 < y0, the exponent is negative. Thisimplies that the integrand evaluated on the semi-circle is vanishing exponentially fast asR → ∞ and the contribution to the integral also vanishes. The same argument applies inthe case where x0 > y0, with the replacement Ep → −Ep since we in this case close thecontour in the lower half-plane

Looking at Eq. (5.76), we conclude that the Fourier transform of DF (x − y) is preciselythe integrand. Denoting this function by DF (p), the propagator in momentum space is

DF (p) =i

p2 −m2 + iǫ. (5.81)

The dispersion relation is then given by the poles of the propagator, Ep =√p2 +m2. This

property is general.

What is the mathematical interpretation of the propagator? It satisfies

[−∂µ∂µ +m2]DF (x− y) =∫

d4p

(2π)4

i(−p2 +m2)

p2 −m2 + iǫe−ip(x−y)

= −i∫ d4p

(2π)4e−ip(x−y)

= −iδ4(x− y) . (5.82)

Hence, the free particle propagator is a Green’s function for the Klein-Gordon operator. Itcreates a particle with momentum p at x with amplitude eipx which propagates to the pointy and is annihilated.

——————————————————————————————————

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CHAPTER 5. FIELD QUANTIZATION 60

Example

We already know the Green’s function in electromagnetism. Consider Maxwell’s equation:

∇ ·E = ρ , (5.83)

where ρ is the charge density. The electric field can be expressed in terms of the scalarpotential Φ as

E = −∇Φ . (5.84)

Combining the two equations, we obtain

−∇2Φ = ρ . (5.85)

For a point source ρ = qδ(r), we find

−∇2Φ = −d2Φ

dr2

= qδ(r) . (5.86)

The function Φ(r) and δ(r) can be written in terms of their Fourier transforms:

Φ(r) =∫

d3p

(2π)3eip·rΦ(p) , (5.87)

δ(r) =∫

d3p

(2π)3eip·r . (5.88)

Inserting these expressions into Eq. (5.86), we obtain

d2

dr2

d3p

(2π)3eip·rΦ(p) = −

d3p

(2π)3eip·rp2 Φ(p)

= −q∫ d3p

(2π)3eip·r . (5.89)

This yields

Φ(p) =q

p2. (5.90)

A straightforward calculation shows that its Fourier transform is

Φ(r) = q∫ d3p

(2π)3e−ip·r

1

p2

=q

4πr. (5.91)

This is simply the Coulomb potential.

——————————————————————————————————

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CHAPTER 5. FIELD QUANTIZATION 61

5.6 Problems

5.6.1 Anharmonic oscillator

Consider the anharmonic oscillator with Hamiltonian

H =p2

2m+

1

2mω2x2 +

λ

24x4 , (5.92)

where λ is a constant.

1) The last term can be considered an interaction, i.e. we write

H = H0 +H1 (5.93)

where

H0 =p2

2m+

1

2mω2x2 , (5.94)

H1 =λ

24x4 . (5.95)

1) Express H1 in terms of a and a†. Consider the terms in H1 with two factors of a and twofactors of a†. Write these terms in normal-ordered form.

2) Use first-order perturbation theory to calculate the change in the energy of the groundstate. Argue that only the terms found in 1) contribute and use this to simplify your calcu-lations.

5.6.2 Momentum operator

Consider the momentum operator for the Klein-Gordon field

P = −∫

d3xπ(x)∇φ(x) . (5.96)

Express this operator as a momentum integral involving ap and a†p in the same way as weexpressed the Hamiltonian.

5.6.3 Nonrelativistic fermions

The Lagrangian for a Schrodinger field ψ is given by

L = −1

2ih(ψ†ψ − ψ†ψ) − h2

2m(∇ψ†)·(∇ψ) . (5.97)

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CHAPTER 5. FIELD QUANTIZATION 62

The Lagrangian is invariant under a global phase transformation

ψ → eiαψ , (5.98)

ψ† → e−iαψ† , (5.99)

where α is a constant. In the following, the field describes bosons.

1) Use the field operator ψ(x, t) to express the conserved charge operator Q that corre-sponds to this symmetry in terms of creation and annihilation operators.

2) Use Heisenberg’s equation of motion

dQ

dt= i[H,Q] , (5.100)

to show explicitly that Q is independent of time.

5.6.4 Integral representation of step function

Prove the integral representation of the step function

θ(t− t′) = −∫ ∞

−∞

2πi

e−iω(t−t′)

ω + iη, (5.101)

where η is a small positive quantity.

5.6.5 Propagator for NR field

Consider a norelativistic field theory with Lagrangian

L = −1

2ih(ψ†ψ − ψ†ψ) − h2

2m(∇ψ†)·(∇ψ) . (5.102)

The equation of motion for ψ is

[

−i∂0 −1

2m∇2]

ψ = 0 . (5.103)

Find the momentum space propagator G(p) by using that the propagator G(x, y) is the

Green’s function of the differential operator[

−i∂0 − 12m

∇2]

.

5.6.6 Propagator for the Dirac field

Repeat the previous problem for relativistic fermions.

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CHAPTER 5. FIELD QUANTIZATION 63

5.6.7 Finite density

In statistical mechanics, we can use different ensembles and their corresponding partitionfunctions. If we consider a system with finite chemical potential, i.e. when the particlenumber is not fixed, we make the replacement

H → H − µN , (5.104)

where H is the Hamiltonian, µ is the chemical potential, and N is the particle number.More generally, we can replace N by any conserved quantity Q (arising from a continuoussymmetry), e .g. electric charge and µ is the chemical potential that corresponds to theconserved quantity.

The replacement can be written as K → H, where

K = H− µj0 , (5.105)

where H is the Hamiltonian density and j0 is the (charge) density.

1) The Lagrangian for relativistic fermions has a global phase symmetry and an associ-ated conserved charge. Introduce a chemical potential for the charge density and use theresult derived earlier to calculate K.

2) Go “backwards” and calculate the Lagrangian density L from the new Hamiltonian densityK. Show that this corresponds to the replacement recipe

∂0 → ∂0 − iµ . (5.106)

Conclude that µ can be regarded as the zeroth component A0 of a gauge field!

3) Repeat the calculations for a complex scalar field. Any comments?

5.6.8 Complex propagator and Greens’ function

Consider the complex scalar field coupled to a chemical potential.

1) Derive the equation of motion for Φ†.

Write the complex scalar field Φ in terms of two real fields, Φ = (φ1 + iφ2)/√

2. Theequation of motion can now be written as

(A + iB)φ1 + (C + iE)φ2 = 0 . (5.107)

2) Determine A,B,C and E.

The equation of motion can be written in matrix form

1√2D−1

(

φ1

φ2

)

= 0 , (5.108)

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CHAPTER 5. FIELD QUANTIZATION 64

where D−1 is a 2 × 2 hermitean matrix where the entries are differential operators. 3) De-termine the matrix D−1.

3) Find the propagator matrix D(p) in momentum space, i. e. find the Greens’ func-tion for the differential operator D−1 in momentum space.

4) The dispersion relation is given by detD−1 = 0, i. e. the zeros of the determinantof the inverse matrix. Any comments? (This generalizes the one-component result that thedispersion relation is given by the poles of the propagator, i. e. the zeros of the inversepropagator).

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Chapter 6

Perturbation theory

In the previous chapter we introduced the propagator, which basically is the correlatorbetween the values of the field φ at two points x and y in spacetime. It is sometimes calledthe two-point correlator. It gives the amplitude of a particle to propagate from x to y.

Richard Feynman 1 introduced a graphical representation of two-point function as well asother n-point functions - the socalled Feynman diagrams. The diagrams can be translatedinto mathematical expressions by a set of rules that can be derived from the path-integralformalism introduced in later courses. Here we will restrict ourselves to a few examples.

The free propagator is represented by a line joining the points x and y in spacetime, seeFig. 6.1.

x y

Figure 6.1: Graphical representation of the free propagator.

6.1 Propagator

In this section, we consider perturbation theory for the interacting field theory whose La-grangian density is given by

L =1

2(∂µφ)(∂µφ) − 1

2m2φ2 − λ

24φ4 . (6.1)

The last term in Eq. (6.1) is an interaction. It is the equivalent in field theory to thex4-term of the anharmonic oscillator discussed in problem 5.6.1. The factor 1/24 = 1/4!is just convention. The Euler-Lagrange equation becomes nonlinear and cannot be solvedexactly. It is therefore impossible to expand the quantum field in terms of a complete set of

1Feynman received the Nobel prize in physics in 1965 jointly with Julian Schwinger and Sin-Itiro Tomon-aga for their fundamental work on QED. He also invented the path-integral formalism.

65

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CHAPTER 6. PERTURBATION THEORY 66

solutions. One approach is to do perturbation theory. i.e. we consider the term λφ4/24 a“small” quantity and carry out a series expansion in λ. This is simply a Taylor expansion inλ and the hope is that it converges (quickly) if λ≪ 1. It should then be sufficient to keep afew terms only in this expansion 2.

x y x yz

Figure 6.2: Graphical representation of the free propagator plus the first loop correction.

We next consider the first correction to the propagator. Pictorially, it is given by theright diagram in Fig. 6.2. A particle is created at x, propagates to z, where it is annihilated.At the point z, another virtual 3 particle is created, propagates back to the point z and isannihilated. At the same point, another particle is created and propagates to the point y.

The first-order correction to the propagator is given by

∼ λ〈0|T

φ(x)φ(y)∫

d4zφ4(z)

|0〉 . (6.2)

After a lot of algebra, one finds a contribution 4

∼ λ∫

d4zDF (x− z)DF (z − z)DF (y − z) . (6.3)

This expression can be obtained from the Feynman diagram by multiplying by −iλ for eachvertex (the point where four lines meet) and DF (v−w) for each propagator line joining thepoints v and w. Inserting the expressions for the propagators, we can write

−iλ2

d4zDF (0)∫ d4p

(2π)4

ie−ip(x−z)

p2 −m2 + iǫ

∫ d4q

(2π)4

ie−iq(z−y)

q2 −m2 + iǫ. (6.4)

Integrating over z gives a delta function, δ4(p− q) and after integrating over q, we can writeEq. (6.4)as

λ

2DF (0)

∫ d4p

(2π)4

ie−ip(x−z)

(p2 −m2 + iǫ)2. (6.5)

Adding this correction to the free propagator, one can write

DF (x− y)int =∫

d4p

(2π)4

ie−ip(x−y)

(p2 −m2 + iǫ)+λ

2DF (0)

d4p

(2π)4

ie−ip(x−y)

(p2 −m2 + iǫ)2, (6.6)

2In QED, the expansion parameter is α ≈ 1/137 ≪ 1 and so perturbative calculations in QED seem tomake sense. However, such series are typically asymptotic series which have zero radius of convergence(!).See e. g. [11].

3A virtual particle does not obey the relation E2 = m2 + p2 as real particles must. They are not on thesocalled mass shell.

4This calculation is somewhat simplified since we haven’t developed the full machinery of QFT.

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CHAPTER 6. PERTURBATION THEORY 67

where DF (x− y)int is a propagator including interactions. This correction can be absorbedin the expression for the free propagator as follows. We write the interacting propagator as

DF (x− y) int =∫

d4p

(2π)4

ie−ip(x−y)

(p2 −m2 − λ2DF (0) + iǫ)

. (6.7)

Expanding the denominator in Eq. (6.7) to first order in the “small” quantity λ, we obtainEq. (6.6) - correct to the order we are working. The pole of the propagator is now given by

p2 −m2 − λ

2DF (0) = 0 . (6.8)

or equivalently the dispersion relation

E2p = p2 +m2 +

λ

2DF (0) . (6.9)

The physical mass is no longer equal to the mass parameter in the Lagrangian. What hashappend??? Due to interactions, the pole of the propagator has moved. In fact, if we takeequation (6.9) at face value, it has become infinte due to quantum fluctuations. The problemis that the correction due to interactions involve virtual particles and quantum fluctuationson all momentum scales since we are integrating over the four-momentum of the particlesfrom zero all the way to infinity. The largest energy scales correspond to the shortest lengthscales. We know that quantum field theory cannot be the correct description of Nature atall length scales. For example, quantum gravity is expected to break down at the Planckscale ∼ 1019 GeV 5. This scale comes about from combining h, c, and Newton’s constant ofgravity GN . So we take a practical approach and say that our theory is valid up to someultraviolet scale Λ. The physics above that scale does not concern us. We then regularize 6

the integral in DF (0) by imposing a cutoff on the radial part of the integral:

DF (0) =∫

d4p

(2π)4

i

p2 −m2 + iǫ

=∫

d3p

(2π)3

1

2Ep

=1

4π2

∫ Λ

0

dp√p2 +m2

=1

8π2

[

Λ√

Λ2 +m2 −m2 logΛ +

√Λ2 +m2

m

]

≈ 1

8π2

[

Λ2 − 1

2m2

(

logΛ2

m2− 1

)]

, (6.10)

5Yes, one can quantize gravity. However, at the length scales probed so far, ordinary gravity, i.e. classicalfield theory is in excellent agreement with observations.

6To regularize an integral means a procedure to make sense of or deal of integrals that are infinite. Thereare many different regularization procedures and one of them is to cut off the integral by integrating to anupper limit Λ. You will learn more about this in FY3464 and FY3466!

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CHAPTER 6. PERTURBATION THEORY 68

where we in the last line have used m ≪ Λ. At this point, we do not know what the UVcutoff scale is, except that it is huge compared to scales we have probed in accelerators sofar. The Large Hadron Collider (LHC) at CERN is constructed for several purposes. Oneof them is to probe smaller length scales than at previous accelerators.

Summarizing our discussion, we can write

m2phys = m2

bare +λ

16π2

[

Λ2 − 1

2m2

(

logΛ2

m2− 1

)]

, (6.11)

where we have renamed the mass parameter m2 of the Lagrangian to m2bare. More generally,

all parameters in the Lagragian are called bare quantities in contrast to the physical ones.They are being dressed by quantum fluctuations. Note that the equation (6.11) that relatesthe bare and the physical mass depends on the scale Λ. Since the physical mass that wemeasure cannot depend on this scale, the bare mass parameter must depend on Λ in sucha manner that mphys does. This implies that if we change the cutoff, we must also adjustthe bare parameters so that physical measured quantities do not. Physically, increasing thecutoff from Λ0 to Λ1, say, means including quantum fluctuations between these two scales.

6.2 Vacuum energy

We next consider the vacuum energy of the φ4-theory. The problem of the vacuum energy hasbeen dubbed as the most serious problem in theoretical physics 7. While the present authorthinks that this is overselling the problem, it does show the difficulties of local quantum fieldtheories. Some clever string theorist will eventually solve it.

The Hamiltonian density is given by

H =1

2

(

∂φ

∂t

)2

+1

2(∇φ)2 +

1

2m2φ2 +

λ

24φ4 . (6.12)

The Hamiltonian density is written as sum of a zeroth-order term and a perturbation ac-cording to H = H0 + H1, where

H0 =1

2

(

∂φ

∂t

)2

+1

2(∇φ)2 +

1

2m2φ2 , (6.13)

H1 =λ

24φ4 . (6.14)

We will use first-order perturbation theory to calculate first-order shift of the vacuum energydensity due to interactions. This yields

E = E0 + E1 , (6.15)

7As shown below, the vacuum energy goes like Λ4, where Λ is the ultraviolet cutoff. This could be thePlanck scale. However, observations of the curvature of the Universe indicates a cosmological constant (orvacuum energy density), which is some 120 orders of magnitude smaller, i.e. an almost flat Universe.

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CHAPTER 6. PERTURBATION THEORY 69

where

E0 = 〈0|H0|0〉 , (6.16)

E1 = 〈0|H1|0〉 . (6.17)

We already know the expression for the zeroth-order term. It is given by the standardexpression

E0 =1

2

d3p

(2π)3Ep . (6.18)

As noted before, this integral is infinite. If we regularize it using an ultraviolet cutoff Λ, weobtain

E0 ≈ 1

16π2

[

Λ4 +m2Λ2 − 1

4m4

(

logΛ2

m2− 1

2

)]

, (6.19)

where ≈ sign is used to indicate that this is the approximate value of the integral for m≪ Λ.We know that the expression for E0 is in terms of m2

bare and we would like to express it interms of the physical mass m2, which differs by a “small” quantity proportional to λ. We dothis by writing m2

bare = m2 − δm2 (skipping the subscript phys) on the physical mass, andsubstituting this into Eq. (6.19). This yields

E0 ≈ 1

16π2

[

Λ4 + (m2 − δm2)Λ2 − 1

4(m2 − δm2)4

(

logΛ2

m2 − δm2− 1

2

)]

≈ 1

16π2

[

Λ4 +m2Λ2 − 1

4m4

(

logΛ2

m2− 1

2

)]

− δm2

16π2

[

Λ2 − 1

2m2

(

logΛ2

m2− 1

)]

,

1

16π2

[

Λ4 +m2Λ2 − 1

4m4

(

logΛ2

m2− 1

2

)]

− λ

4[DF (0)]2 , (6.20)

where we in the last second line has expanded to first order in δm2 i. e. kept all termsthrough order λ, and used that δm2 = λDF (0)/2.

Let us now look at the first-order shift given by

E1 =λ

24〈0|φ4(x)|0〉 . (6.21)

Note that one might think that the vacuum energy density is a function of x due to φ(x).However, since the vacuum is Lorentz invariant, it turns out to be a constant in space andtime, as we shall see below. The perturbation is given by products of four field operatorsand can be written as

φ4(x) =∫

d3p

(2π)3

d3q

(2π)3

d3r

(2π)3

d3s

(2π)3

1

2Ep2Eq2Er2Es[

apa†qara

†se

−i(p−q+r−s)x + ...]

, (6.22)

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CHAPTER 6. PERTURBATION THEORY 70

where the ellipsis denote the remaining 15 terms you get by multiplying out φ4(x) in productsof creation and annihilation operators. The trick is now to normal order all the interactionterms just like we did for the anharmonic oscillator (problem 5.6.1). For example, considerthe term apa

†qara

†s. Using the commutation relations, we can write

apa†qara

†s =

[

(2π)3δ3(p− q) + a†qap] [

(2π)3δ3(r − s) + a†sar]

= (2π)6δ3(p− q)δ3(r − s) + a†qapa†sar + ... (6.23)

Now all the operator products in Eq. (6.23) as well as the other terms in φ4(x) will eventuallybe normal ordered. They will therefore annihilate the vacuum state |0〉. The only termsthat contribute are the constants such as the first term in Eq. (6.23) There are three suchidentical terms (cf. the factor three in problem 5.6.1), and so we can write

E1 =λ

8

∫ d3p

(2π)3

∫ d3q

(2π)3

∫ d3r

(2π)3

∫ d3s

(2π)3

1

2Ep2Eq2Er2Es

×(2π)6δ3(p− q)δ3(r − s)e−i(p−q+r−s)x . (6.24)

Integrating over q and s (and letting r → q in one of the integrals) yields

E1 =λ

8

d3p

(2π)3

d3q

(2π)3

1

2Ep2Eq. (6.25)

Recalling that the propagator DF (x) can be written as

DF (x) =∫

d3p

(2π)3

1

2Epe−ipx , (6.26)

one finds

E1 =λ

8[DF (0)]2 . (6.27)

Note that we do not distinguish between the bare and the physical mass in Eq. 6.27) sinceit is already proportional to λ. The difference will be of higher order in λ. If we were to goto second order in perturbation theory, we would have to be careful to calculate consistentlyto order λ2. This would also include the order-λ2 correction to the physical mass.

If we again represent a propagator DF (x− y) by a line with endpoints x and y, the first-order shift can be represented by two bubbles starting and ending at the same point. This isanother example of a Feynman diagram. This is shown in Fig. 6.3. Since there are no openlines, it is called a vacuum diagram.

Adding Eqs. (6.19) and (6.27), we obtain the full first-order result in vacuum energyexpressed in terms of the physical mass.

E =1

16π2

[

Λ4 +m2Λ2 − 1

4m4

(

logΛ2

m2− 1

2

)]

− λ

8[DF (0)]2

= .1

16π2

[

Λ4 +m2Λ2 − 1

4m4

(

logΛ2

m2− 1

2

)]

− λ

8

1

64π4

[

Λ2 − 1

2m2

(

logΛ2

m2− 1

)]2

.

(6.28)

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CHAPTER 6. PERTURBATION THEORY 71

x

Figure 6.3: Vacuum diagram: two-loop correction to the ground-state energy.

6.3 Dimensional regularization

Calculating the quantum correction to the mass from vacuum fluctuations, we found

δm2 =λ

16π2

[

Λ2 − 1

2m2

(

logΛ2

m2− 1

)]

.

In the limit Λ → ∞, the result is divergent. Moreover there are two types of divergences.There are power divergences and logarithimic divergences. One might argue that the latterare more physical than the former, since a logarithmic divergences always has to match thelogarithm of another scale giving rise to the logarithm of a dimensionless quantity 8.

There is another and generally more convenient and elegant way to regularize divergentintegrals and the method is called dimensional regularization 9 invented by ’t Hooft andVeltman in 1972 [10] 10. Consider the integral

Idn =∫

ddp

(2π)d1

(√p2 +m2)n

, (6.29)

where d is the dimension of space. For n = 2 this integral is divergent in the ultraviolet ininteger dimensions d = 2, 3..., but convergent in d = 1. In d = 2 it logarithmically divergent,in d = 3 it is linearly divergent and so forth:

I22 ≈ 1

2m, (6.30)

I22 ≈ 1

4πlog

Λ2

m2, (6.31)

I32 ≈ 1

2π2

[

Λ − π

2m]

, (6.32)

where we again have used m≪ Λ.

8The logarithm of a dimensionful number is in itself meaningless.9Dimensional regularization respects gauge invariance and Lorentz invariance. The three-dimensional

cutoff Λ does not.10It is often called ”dimreg” by practitioners.

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CHAPTER 6. PERTURBATION THEORY 72

Assume now that we define it any dimension by analytic continuation. For example, if weare interested in the integral in d = 2 dimensions, we write d = 2 − 2ǫ, where ǫ is a positivenumber. The integral is then defined by

Idn = µ2−d∫ ddp

(2π)d1

(√p2 +m2)n

. (6.33)

Here we have introduced a new mass scale µ so that the integral has mass dimension two, alsowhen d 6= 2. The first thing we need to do, is to define the angular integral in d dimensions.For d = 1, 2, 3..., the “area” of the sphere Sd is given by

Sd =2π

d2

Γ(d2), (6.34)

where Γ(x) is the Gamma function. We then define the area of the sphere in nonintegerdimensions d by Eq. (6.34). The remaining radial integral needs to be generalized to ddimensions such that it reduces to the standard expression for d = 1, 2, 3. It can thereforebe defined as

Idn(radial) =∫ ∞

0

pd−1dp

(√p2 +m2)n

. (6.35)

A straightforward calculation yields

Idn(radial) =m(d−n)/2Γ(d

2)

2Γ(n2)

Γ(

n−d2

)

. (6.36)

Putting the pieces together, one finds

Idn =(

µ

m

)2−d m2− d2−n

2 Γ(n−d2

)

2dπd2 Γ(n

2)

. (6.37)

We next expand our result in powers of ǫ i. e. we expand around d = 2 dimensions. Forexample, for n = 2 this yields

Id2 =1

[

1

ǫ+ log

µ2

m2+ constant + O(ǫ)

]

. (6.38)

We first notice that there is a pole in ǫ 11 and a logarithmic term associated with. If wecompare this result with the one using a cutoff Λ, we notice that the logarithms are thesame except that we have traded Λ for the scale µ. This is always the case. If an integral islogarithmically divergent using a cutoff, there is pole and a corresponding logarithmic termif one uses dimensional regularization.

11Note that the Γ(n)-function has poles at n = 0,−1,−2....

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CHAPTER 6. PERTURBATION THEORY 73

Returning to the orginal integral Id1 = 2DF (0) for d = 3 − 2ǫ, one finds 12

Id1 =(

µ

m

)3−d m4− d2− 1

2 Γ(1−d2

)

2dπd2 Γ(1

2)

= −m2

8π2

[

1

ǫ+ log

µ2

m2+ constant + O(ǫ)

]

. (6.39)

Again we see that the logarithm matches the original one in Eq. (6.10). Moreover, the powerdivergence has disappeared. This is also always the case: Power divergences are set to zerowith dimensional regularization. If an integral has only such divergences, it will be finitewith dimensional regularization. This is hexerei 13.

Using the expression (6.39), we obtain

m2 = m2bare −

λm2bare

32π2

[

1

ǫ+ log

µ2

m2

]

+ δm2 , (6.40)

where we have added a socalled counterterm δm2, and where we have ignored the constant,which is immaterial for our discussion. The counterterm is now chosen such that the physicalmass of the particle is finite. The simplest choice is

δm2 =λm2

bare

32π2

1

ǫ, (6.41)

i.e. we simply use it to cancel the pole in ǫ. The physical mass then reads

m2 = m2bare +

λm2bare

32π2log

µ2

m2bare

. (6.42)

The procedure of rendering quantities finite by adding appropriate counterterms is calledrenormalization and is necessary in most field theories renormalization 14. The reason behindthese divergences is that interactions in spacetime take place at a single point and is thereforehighly singular. String theory is an attempt to tame the divergences via interactions thatare “blurred”. A beatiful and deep article on renormalization was written by Lepage andcan be found at [12]. It is an excellent read.

6.3.1 Vacuum energy revisited

The zeroth-order vacuum energy density is given by

E0 =1

2µ3−d

ddp

(2π)3

p2 +m2 . (6.43)

12This integral is divergent in all integer dimension. However, it is still defined the same way in dimensionalregularization. Just multiply with the appropriate factor of µ.

13Hexerei also implies that one must be careful with applying dimreg carelessly. There are a few pitfalls.14Quantum mechanics can be considered a field theory in 0+1 dimensions but is finite.

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CHAPTER 6. PERTURBATION THEORY 74

With dimensional regularization this reads

E0 = m4(

µ

m

)2ǫ Γ(n−d2

)

2d+1πd/2Γ(12). (6.44)

Expanding in powers of ǫ, one obtains

E0 = − m4

64π2

[

1

ǫ+ log

µ2

m2+ C + O(ǫ)

]

. (6.45)

Again we see that the power divergences are set to zero and that we need a counterterm

δE0 =m4

64π2ǫ. (6.46)

to cancel the divergences in the vacuum energy. Thus the renormalized vacuum energy is

E0 = − m4

64π2log

µ2

m2+ C . (6.47)

The first-order correction to the ground-state energy can likewise be written as

E1 =λ

8[D(0)]2

8

m4

256π4

[

1

ǫ2+

2

ǫlog

µ2

m2+ log2 µ

2

m2+K

]

+ δE1 . (6.48)

Expressing the zeroth-order energy density in terms of the physical mass instead of the baremass is equivalent to the replacement E0 → E0 + δm2

2DF (0) 15. This yields

E = E0 +δm2

2DF (0) +

λ

8[DF (0)]2

= − m4

64π2log

µ2

m2− λ

8

m4

256π4

[

1

ǫ2− log2 µ

2

m2+ ...

]

+ δE1 . (6.49)

Note that the term 2ǫlog µ2

m2 in Eq. (6.48) is cancelled by a similar term coming fromδm2

2DF (0). The remaining divergence, which is a double pole in ǫ, is cancelled by the vacuum

counterterm

δE1 =λ

8

m4

256π4ǫ. (6.50)

The renormalized vacuum energy is then

E = − m4

64π2

[

logµ2

m2+ C

]

8

m4

256π4

[

log2 µ2

m2+K1

]

. (6.51)

15Perhaps the easiest way to see this, is to realize that the counterterm gives rise to an extra term in theLagrangian, δL = 1

2δm2φ2 and this terms should be treated as a perturbation since it is of order λ. It then

gives a correction to the vacuum energy δE = δm2

2DF (0).

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CHAPTER 6. PERTURBATION THEORY 75

6.4 Problems

6.4.1 Dimensional regularization

In the theory of Bose-Einstein condensation, the dispersion relation for quasiparticles is givenby

Ep =

p2

2m

(

p2

2m+ 2µ

)

, (6.52)

where m is the mass of the particle and µ is the chemical potential.

1) What is the behavior of Ep for large values of p? And for small values of p?

2) The ground-state energy of the system is then

E =1

2

ddp

(2π)dEp + constant , (6.53)

where the constant is of no interest to us. Calculate the integral in d = 1, 2, and 3 dimensionsusing a cutoff Λ. Analyze the divergences.

3) Repeat the calculation using dimensional regularization. You might want to use Mathe-matica, Maple, or Matlab to calculate the radial integral. Compare the poles in ǫ with thecorresponding divergences using a cutoff Λ.

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Chapter 7

Dilute Bose gas

7.1 Lagrangian and Hamiltonian

The starting point for the discussion on the dilute Bose gas the Lagrangian density

L = iψ†∂tψ + µψ†ψ − 1

2m∇ψ† · ∇ψ − 1

2g(

ψ†ψ)2

, (7.1)

The last term is an interaction terms, which describes two-particle scattering. The couplingconstant g is basically the s-wave scattering length Note that the Lagrangian is invariantunder the global phase symmetry ψ → eiαψ even in the presence of interactions. This givesrise to a conserved current in the usual way. In the nonrelativistic case, it is simply numberconservation (of atoms) and the number density is given by j0 = ρ = 〈ψ†ψ〉. In orderto couple the theory to a a chemical potential, we have therefore made the substitution∂0 → ∂0 − iµ. The Hamiltonian density is then 1

K =1

2m∇ψ† · ∇ψ − µψ†ψ +

1

2g(

ψ†ψ)2

. (7.2)

We then write

K = H− µψ†ψ , (7.3)

where we have defined

H =1

2m∇ψ† · ∇ψ +

1

2g(

ψ†ψ)2

. (7.4)

The integrals of K and H are denoted by

K =∫

d3xK , (7.5)

H =∫

d3xH . (7.6)

1When we include a chemical potential, it is often denoted by K, sometimes called the ”Kamiltonian”density.

76

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CHAPTER 7. DILUTE BOSE GAS 77

Taking the derivative of K with respect to µ, we obtain

∂K

∂µ= −

d3xψ†ψ

= −〈N〉 . (7.7)

The expectation value of K is by definition the Helmholtz free energy density F and so wefind

F = 〈K〉= 〈H − µψ†ψ〉= 〈H〉 − µ〈ψ†ψ〉= E − µρ (7.8)

where the expectation value of the energy density is E = 〈H〉 and the expectation value ofthe number density is ρ = 〈ψ†ψ〉. This yields

E = F + µρ . (7.9)

The free energy density is a Legendre transform of the energy density.

Consider the ground state of an ideal Bose gas with N0 particles in a volume V . Theparticle density is therefore ρ0 = N0/V . The thermodynamic limit is defined by taking thelimit N0, V → ∞ with fixed density ρ0. Since the particles are bosons, they all recide in thesingle-particle ground state with p = 0. This state is denoted by |Φ〉 and is given by

|Φ〉 =(

a†0)N0 |0〉 . (7.10)

Let us calculate the charge density I in the ground state

I = 〈Φ|Q|Φ〉 , (7.11)

where Q is the ”charge” density operator that arises from the global phase symmetry

Q =1

V

p

a†pap , (7.12)

where we for convenience have switched to the discrete case summing over all values of themomentum p. This is just to simplify intermediate calculations. We will take the continuumlimit at the end. This yields

I =1

V

p

〈Φ|a†pap|Φ〉

=1

V

p

〈Φ|a†pap

(

a†0)N

0|0〉

=1

V

p

〈Φ|a†p[

a†0ap + δp,0] (

a†0)N0−1 |0〉 ,

=1

V〈Φ|Φ〉 +

1

V

p

〈Φ|a†pa†0ap

(

a†0)N0−1 |Φ〉 .

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CHAPTER 7. DILUTE BOSE GAS 78

We can now us the commutation relation to move ap one more step to the right. Every timewe do it, we pick up a matrix element 〈Φ|Φ〉/V . Doing it N0 times, it brings ap to the right

of all the a†0’s and gives zero since it annihilates the vacuum. We are then left with

I =N0

V〈Φ|Φ〉 . (7.13)

Since the ground state is assumed to have unit norm, this reduces to the density ρ = N0/Vof the ground state as expected:

I = ρ , (7.14)

In the same manner we can calculate the matrix element

J =1

V

p

〈Φ|apa†p|Φ〉 . (7.15)

Using the commutation relations, it is not difficult to show that

J =(N0 + 1)

V〈Φ|Φ〉 . (7.16)

When N0 is a macroscopic number, N0 ≫ 1, we can therefore ignore the commutator betweena0 and a†0 and simply treat the operators as numbers 2. This is exact in the thermodynamiclimit when N0 → ∞. The replacement of the operators a0 and a†0 by

√ρ is called the

Bogliubov replacement and is the basis of the Bogoliubov approximation which we discussnext.

7.2 Bogoliubov approximation

We next rewrite the field operators by separating out the p = 0 mode from the integral

ψ(x, t) =√ρ+

1√V

′∑

p

ape−ipx , (7.17)

ψ†(x, t) =√ρ+

1√V

′∑

p

a†peipx , (7.18)

where the prime on the sum is a reminder that p = 0 is excluded and V is the volume ofthe system Inserting the expression for the field operator, we can write

H = H0 +Hint , (7.19)

2By neglecting the operator nature of a0 and a†0, we treat them as classical variables. We refer to themas c-numbers.

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CHAPTER 7. DILUTE BOSE GAS 79

where

H0 =1

V

d3x′∑

p,q

p · q2m

a†paqei(p−q)x , (7.20)

Hint =1

2g∫

d3xN2

0

V 2+gρ0

2V

d3x′∑

p,q

[

a†pa†qei(p+q)x

+apaqe−i(p+q)x + a†paqe

i(p−q)x + ...]

(7.21)

where the dots indicate the remaining terms, for example terms with more than two operatorswith nonzero momenta, e.g.

√ρa†paqake

i(p−q−k)x. Since we expect most particles to be inp = 0 state also in the interacting case, we will neglect the terms with more than twooperators with nonzero momenta 3. These terms can be viewed as perturbations and canbe taken systematically into account using perturbation theory. The first term in Eq. (7.21)is rewritten as follows

1

2g∫

d3xN2

0

V 2=

1

2g∫

d3x1

V 2[N −N ′]2

≈ 1

2g∫

d3x1

V 2[N2 − 2NN ′] ,

=1

2gV ρ2 − gρN ′

=1

2gV ρ2 − gρ

′∑

p

a†pap , (7.22)

where N ′ is the number of particles outside the condensate. Note that we have dropped theterm (N ′)2 as it is expected to be small compared to the terms we kept.

After integrating over x and then summing over q, the second term in Eq. (7.21) reducesto (cf. Eq. (A.13))

gρ0

2V

d3x′∑

p,q

[

a†pa†qe

i(p+q)x + apaqe−i(p+q)x + a†paqe

i(p−q)x + ...]

=gρ

2

′∑

p

[

a†pa†−p + apa−p + 4a†pap

]

. (7.23)

where we have replaced ρ0 by ρ, which is correct to the order we are calculating. AddingEqs. (7.20), (7.22) and (7.23) , we obtain

H =1

2gV ρ2 +

′∑

p

p2

2ma†pap +

2

′∑

p

[

a†pa†−p + apa−p + 2a†pap

]

. (7.24)

We next divide the Hamiltonian (7.24) to obtain the Hamiltonian density

H =1

2gρ2 +

1

V

′∑

p

p2

2ma†pap +

2V

′∑

p

[

a†pa†−p + apa−p + 2a†pap

]

. (7.25)

3Interactions are supposed to be weak! (whatever that means at this stage).

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CHAPTER 7. DILUTE BOSE GAS 80

We have now reduced the problem to that of a Hamiltonian density which is quadratic inthe creation and annihilation operators and is essentially that of a free theory. However, wehave some troublesome terms containing creation and annihilation operator. The next stepis to introduce new operators bp and b†p expressed in term of the old ones such that theysatisfy the standard commutation relations and such that the Hamiltonian is ”diagonal” inthese operators. The problem is then reduced to problem we know how to solve. To thisend, we define

a†p = upb†p + vpb−p (7.26)

ap = upbp + vpb†−p . (7.27)

This transformation is called a Bogoliubov transformation after the famous Soviet physicistN. N. Bogoliubov. Note that the coefficients up and vp are chosen real and sphericallysymmetric. This choice is without loss of generality. These equations are easily solved withrespect to the new operators

bp =1

v2p − u2

p

[

vpa†−p − upap

]

, (7.28)

b†p =1

v2p − u2

p

[

vpa−p − upa†p

]

. (7.29)

Inserting these defintions into the commutator, we find

[bp, b†q] =

1

u2p − v2

p

1

u2q − v2

q

[

upuq[ap, a†q] + vpvq[a

†−p, a−q]

]

=1

u2p − v2

p

1

u2q − v2

q

[upuq − vpvq] δp,q . (7.30)

We now require that the new operators satisfy the standard commutation relation for bosonicoperators. This implies

u2p − v2

p = 1 , (7.31)

for all p 6= 0. We can parametrize up and v−p as follows

up = coshαp , vp = sinhαp , (7.32)

where αp is a real number. We next insert the expressions for ap and a†p into the Hamiltonianand collect the various terms. After some tedious algebra, we find

H =1

2gρ2 +

1

V

′∑

p

[(

p2

2m+ gρ

)

(

u2p + v2

p

)

+ 2gρupvp

]

b†pbp

+

[(

p2

2m+ gρ

)

upvp +1

2(u2

p + v2p)gρ

]

[

b†pb†−p + bpb−p

]

+

[(

p2

2m+ gρ

)

v2p + 2gρupvp

]

. (7.33)

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CHAPTER 7. DILUTE BOSE GAS 81

In order to diagonalize the Hamiltonian, we require that the coefficients of b†pb†−p and bpb−p

vanish, i.e. we require

[

p2

2m+ gρ

]

upvp +1

2gρ[

u2p + v2

p

]

= 0 . (7.34)

Using the fact that u2p = 1 + v2

p, we can eliminate vp, and obtain a quartic equation for up

[

p2

2m+ gρ

]2

u2p(u

2p − 1) − 1

4g2ρ2

[

2u2p − 1

]2= 0 . (7.35)

The solution to this equation is easily found and we finally obtain for up and vp

u2p =

1

2

(

p2/2m+ gρ

ǫ(p)+ 1

)

, (7.36)

v2p =

1

2

(

p2/2m+ gρ

ǫ(p)− 1

)

, (7.37)

where we the dispersion relation is

ǫ(p) =

p2

2m

(

p2

2m+ 2gρ

)

. (7.38)

The spectrum (7.38) was first obtained by Bogoliubov [14]. and is referred to as the Bogoli-ubov spectrum. Inserting the expressions for up and vp into the Hamiltonian density (7.33)can now be written as

H =1

2gρ2 +

1

2V

′∑

p

[

ǫ(p) − p2

2m− gρ

]

+1

V

′∑

p

ǫ(p)b†pbp . (7.39)

We have transformed the Hamiltonian density into one which is the sum of two constant termsplus a term which is quadratic in the creation and annihilation operators. These operatorssatisfy the standard commutation relation and we therefore have a particle interpretation ofthe theory. These particles are referred to as quasiparticles and are not reminiscent of thoseof the free theory. Their properties depend on the medium. The first term is a socalledmean-field term and is the classical contribution to the energy density Moreover, the secondterm is a divergent contribution to the energy density. It is due to quantum fluctuationsand is the analog of the zero-point energy hω/2 of the harmonic oscillator. However, sinceit involves the coupling constant, it requires renormalization.

We denote the ground state state by |Φ〉 and it is defined by the absence of quasiparticles,i. e. 4

bp|Φ〉 = 0 . (7.40)

4The absence of quasiparticles is not the same as absence of real particles.

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CHAPTER 7. DILUTE BOSE GAS 82

We next discuss the Bogoliubov dispersion relation (7.38). At long wavelengths, p2/2m≪ gρ,we see that

ǫ(p) ≈√

mp

= cp , (7.41)

where c =√

gρ/m. The dispersion relation is linear at long wavelengths. These are calledphonons or sound waves. The dispersion relation for small momenta is very different fromthat of free particles, ǫ(p) = p2/2m and is due to medium effects. For large momenta, wep2/2m≫ gρ, we see that

ǫ(p) ≈ p2

2m+ gρ .

This is the dispersion relation of a free particle with an additional mean-field energy gρ. Thedispersion relation normalized to gρ is shown in Fig.7.1.

0

2

4

6

8

10

12

14

0 1 2 3 4 5pcgρ

ǫ(p)/gρpc/gρ

(pc)2/2(gρ)2 + 1

Figure 7.1: The dispersion relation normalized to gρ as well as the small - and large-p limitsas functions of cp/gρ.

The average particle number outside the condensate is given by the integral

N ′ =′∑

p

〈a†pap〉 , (7.42)

where

〈a†pap〉 = 〈Φ|(upb†p + vpb−p)(upbp + vpb†−p)|Φ〉

= v2p . (7.43)

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CHAPTER 7. DILUTE BOSE GAS 83

This yields

ρ′ =1

V

′∑

p

v2p

=∫

d3p

(2π)3v2p

=1

2

d3p

(2π)3

[

p2/2m+ gρ

ǫ(p)− 1

]

, (7.44)

where we in the second line has taken the thermodynamic limit, i.e. V → ∞, N → ∞ suchthat the density of the system is constant. The large-p behavior of ǫ(p) is given by

ǫ(p) =p2

2m+ gρ− 1

8

g2ρ2

p2/2m+ ... (7.45)

From the large-p behavior of ǫ(p), it follows that the integal in Eq. (7.44) is finite. Astraightforward calculation gives

ρ′ =8

3√π

(

mgρ

)3/2

=8

3

ρ3a3

π. (7.46)

The total density then becomes

ρ = ρ0 + ρ′

= ρ0

1 +8

3

ρ0a3

π

. (7.47)

Note that we have replaced ρ by ρ0 in the second term as the difference is supposed to besmall. The fact that not all particles is in the p = 0 state is a quantum effect. A fraction ofthe particles are kicked out of the ground state and the effect is is referred to as depletion.The result was first derived by Bogoliubov in 1947 [14].

The ground state is given by the state with no quasiparticles present. The energy of theground state is denoted by E0 and the corresponding energy density by E . It follows fromthe Hamiltonian (7.39) that the latter is given by

E =1

2gρ2 +

1

2V

′∑

p

[

ǫ(p) − p2

2m− gρ

]

=1

2gρ2 +

1

2

d3p

(2π)3

[

ǫ(p) − p2

2m− gρ

]

. (7.48)

It follows from Eq. (7.45) that the above integral is ultraviolet divergent. The divergentterm is given by

Ediv = −1

2mg2ρ2

∫ d3p

(2π)3

1

p2. (7.49)

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CHAPTER 7. DILUTE BOSE GAS 84

Having isolated the divergence, the energy density can be rewritten as

E =1

2gρ2

[

1 −mg∫ d3p

(2π)3

1

p2

]

+1

2

∫ d3p

(2π)3

[

ǫ(p) − p2

2m− gρ+

mg2ρ2

p2

]

. (7.50)

The first integral is the same as the expression for the scattering length in the second Bornapproximation! We then redefine the coupling constant by replacing the first term by thephysical, i. e. the measured coupling constant g 5. The second integral is finite since wehave subtracted the divergent piece. This yields

E =1

2gρ2 +

1

2

d3p

(2π)3

[

ǫ(p) − p2

2m− gρ+

mg2ρ2

p2

]

. (7.51)

After having performed the integral, the ground-state energy density can then be written as

E =2πaρ2

m

1 +128

15

ρa3

π

. (7.52)

The first term is the mean-field energy while the second term is the first quantum correctionto the energy density. We notice that the dimensionless expansion parameter for the ground-state energy is the gas parameter

√ρa3. The result (7.52) was first obtained by Lee, Huang,

and Yang [15, 16]. The weakly interacting Bose gas is then defined by√ρa3 ≪ 1. In

this case, one hopes that the first correction in Eq. (7.52) will be small compared to the

mean-field result E0 = 2πaρ2

m. By Legendre transforming, one can compute the free energy

density and therefore the pressure since P = −F . By measuring the s-wave scatteringlength a in scattering experiments, we have all the parameters in the equation of state.By measuring the pressure for various densities, we should be able to see the effects of thequantum fluctuations, i. e. the deviations from mean-field theory!

7.3 Bose-Einstein condensation and spontaneous sym-

metry breaking

The field operator is written as

ψ(x, t) =

N0

V+

1√V

′∑

p

ape−ipx . (7.53)

What is the expectation value of ψ in the interacting ground state? In the trivial groundstate, i.e. in the vacuum, the expectation value is zero since the annihilation operator ap

by definition annihilates |0〉. We know that the interacting nontrivial ground state |Φ〉 is

5We define g = gbare + δg. The counterterm δg then cancels the divergence in the usual way.

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CHAPTER 7. DILUTE BOSE GAS 85

a complicated sum of states with different particle numbers (recall the fluctuations of theoperator Np). One finds

〈Φ|ψ|Φ〉 =

N0

V+

1√V

′∑

p

〈Φ|ape−ipx|Φ〉 . (7.54)

The first term on the right-hand-side in nonzero, while the second can be written as

1√V

′∑

p

〈Φ|ape−ipx|Φ〉 =

1√V

′∑

p

〈Φ|[upbp + vpb†−p]e−ipx|Φ〉

= 0 , (7.55)

since the quasiparticle operator bp and b†−p annihilate the states |Φ〉 and 〈Φ|, respectively.We can therefore write

〈Φ|ψ|Φ〉 =

N0

V≡ φ0 , (7.56)

so φ0 is the square root of the condensate density ρ0. This yields

ψ = φ0 + ψ , (7.57)

where φ0 is a classical field and ψ is a quantum fluctuating field 6. The expectation value ofthe fluctuating field is zero. We next insert the expression for the field operator (7.57) intothe Kamiltonian density. This yields

K =1

2m∇ψ† · ∇ψ − µ

[

φ20 + φ0ψ + φ0ψ

† + ψ†ψ]

+1

2g[

φ20 + φ0ψ + φ0ψ

† + ψ†ψ]2.

(7.58)

Neglecting the quantum fluctuations in the Kamiltonian yields

K = −µφ20 +

1

2gφ4

0 (7.59)

K =∫

d3x[

−µφ20 +

1

2gφ4

0

]

. (7.60)

We define the effective potential 7 by V =[

−µφ20 + 1

2gφ4

0

]

. The value that minimizes theeffective potential, i. e. which gives the ground-state energy depend on the sign of µ. Theminimum is given by

V ′ = −µφ0 + gφ30

= 0 . (7.61)

6It is called a classical field since we are ignoring the commutatator [a0, a†0].

7At this level, V the effective potential is equal to the classical potential V0 in analogy to the anharmonicoscillator in quantum mechanics. There are, however, quantum corrections to it.

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CHAPTER 7. DILUTE BOSE GAS 86

If µ < 0, the minimum is given by φ0 = 0, while for µ > 0, one finds that φ0 is

φ20 = µ/g . (7.62)

Inserting this value into the effective potential V and recalling that φ20 = ρ0, we obtain

V = −µ2

2g. (7.63)

In other words 〈K〉 = −µ2

2g. This yields

〈H〉 = 〈K〉 + µρ0

=1

2gρ2

0 , (7.64)

where we have used ρ0 = −∂〈K∂µ

= µ/g and expressed 〈H〉 in terms of ρ0 instead of µ (Legendre

transform). So far we have chosen φ0 real, but we can equally well consider have chosen itwith a nonzero phase (recall ψ is a complex field). The potential is then written as

V = −µ|φ0|2 +1

2g|φ0|4 . (7.65)

The potential is invariant under global phase transformations. It has the form of a Mexicanhat - see Fig. 7.2.

Figure 7.2: Plot of potential V as a function of magnitude and phase.

The fact that the system picks out a phase, implies that the ground state is no longerinvariant under the global phase transformation - it breaks the global symmetry 8 This isnot as scary as it sounds. Actually, it is analogous to what is happening in a ferromagnet

8This is related to the fact that the ground state is not an eigenstate of the number operator.

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CHAPTER 7. DILUTE BOSE GAS 87

as you cool it below a critical temperature and it acquires a nonzero magnetization. TheHamiltonian is invariant under spatial rotations, but the systems picks out a a direction -the direction of the magnetization. The ground state then breaks the rotational symmetryof the Lagrangian.

7.4 Problems

7.4.1 Complex scalar field, symmetry breaking, and Goldstone

mode

Consider a complex scalar field with an quartic interaction

L = (∂µΦ)∗(∂µΦ) −m2Φ∗Φ − λ

6(Φ∗Φ)2 . (7.66)

The Lagrangian is invariant under global phase transformation. The complex field can bewritten in terms of two real fields φ1 and φ2

Φ =1√2(φ1 + iφ2) . (7.67)

Assume that φ1 has a nonzero vacuum expectation value v in analogy with the dilute Bosegas 9. We therefore write

Φ =1√2(v + φ1 + iφ2) , (7.68)

where v is a classical field and φ1 and φ2 are quantum fluctuating fields.

1) Substitute Eq. (7.68) into the Lagrangian (7.66) and expand it to second order in thefluctuating fields i. e. neglect terms that are third order or higher in φ1 and φ2.

2) Find the classical potential V and show that v 6= 0 at its minimum if m2 < 0.

3) Find the terms that are linear in the fluctuating fields and show that they vanish ifthey are evaluated at the minimum of V .

4) Find the dispersion relations for the fields and show that there is a Goldstone boson.

7.4.2 Weakly interacting Bose gas

Consider the weakly interacting Bose gas.

9In this case, the chemical potential is zero, so we cannot interpret it as a condensate of particles in thep = 0 state. It is more like the spontaneous magnetization in a ferromagnet.

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CHAPTER 7. DILUTE BOSE GAS 88

1) Calculate the fluctuations of the particle number 〈(∆Np)2〉 in a state with momentum p,i.e. calculate

〈N2p〉 − 〈Np〉2 , (7.69)

where Np = a†pap. Take the limits p→ 0 and p→ ∞. Comments? Calculate the integral

I =∫

d3p

(2π)3〈(∆Np)2〉 . (7.70)

2) The ground-state energy density of the Bose gas is

E =2πaρ2

m

1 +128

15

ρa3

π

. (7.71)

Calculate µ, which is given by

µ =dEdρ

. (7.72)

What is the (dimensionless) expansion parameter of µ? Calculate the free energy densityF = E − µρ. Recall F is a function of µ and not ρ!

3) Define

φ(x, t) =1√V

′∑

p

ape−ipx . (7.73)

Calculate

J(|x − y|, t− t′) = 〈Φ|T

φ(x, t)φ†(y, t′)

|Φ〉 , (7.74)

where Φ is the ground state of the weakly interacting Bose gas. Find its Fourier transformJ(p).

7.4.3 Dimensional regularization

The energy of a weakly interacting Bose gas is given by

E =1

2gρ2 +

1

2

∫ d3p

(2π)3

[

ǫ(p) − p2

2m− gρ

]

. (7.75)

1) Calculate separately the three integrals in the above expression using dimensional regu-larization. Comments?

2) Use the results in 1) to rederive Eq. (7.52). Why do you need no counterterms in “dimreg”to render the ground-state energy finite?

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Chapter 8

Fermi gases

We will now take a closer look at Fermi gases. We first consider an ideal Fermi gas at finitechemical potential. Then we will introduce interactions and compute quantum correctionsto the various physical quantities.

8.1 Perfect Fermi gas

The field operator ψ(x) can be expanded as

ψ(x) =1√V

p,λ

ap,λχ(λ)e−ipx , (8.1)

where λ denotes other quantum numbers, e. g. that are necessary to completely specifythe state, and χ(λ) is the associated part of the wavefunction. For example, for a spin-1

2

particle, we may write

χ(λ = 12) =

(

10

)

, χ(λ = -12) =

(

01

)

. (8.2)

The Fermi propagator iG(x, y) is defined by

iG(x, y) = 〈Φ|Tψ(x)ψ†(y)|Φ〉 (8.3)

=

〈Φ|ψ(x)ψ†(y)|Φ〉 , y0 < x0 ,−〈Φ|ψ†(y)ψ(x)|Φ〉 , x0 < y0 ,

(8.4)

where T denotes time-ordering as usual. Note the minus sign when x0 < y0, which is notincluded in the definition of the bosonic propagator (see chapter six). We first consider thecase y0 < x0. Inserting the expansion Eq. (8.1) of the field operator into the first line ofEq. (8.4), we obtain

iG(x, y) = θ(x0 − y0)∑

λ

d3p

(2π)3

d3q

(2π)3〈Φ|e−i(px−qy)ap,λa

†q,λ|Φ〉 . (8.5)

89

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CHAPTER 8. FERMI GASES 90

Note that we have taken the continuum limit and used the orthogonality between the func-tions, χ(λ1)

∗χ(λ2) = δλ1,λ2. Using the anticommutator, we can write this as

iG(x, y) = θ(x0 − y0)∑

λ

d3p

(2π)3

d3q

(2π)3e−i(px−qy)〈Φ|(2π)3δ3(p − q) − a†q,λap,λ|Φ〉 .(8.6)

In the fermionic ground state, we know that all the states up to Fermi energy ǫF are occupiedand all other states are empty. Thus p must be smaller than the Fermi momentum 1 pF ,when ap,λ acts on |Φ〉. To recover the state |Φ〉 when aq,λ acts afterwards, we must haveq = p. This yields

iG(x, y) = θ(x0 − y0)∑

λ

d3p

(2π)3

d3q

(2π)3e−i(px−qy)〈Φ|(2π)3δ3(p − q)(1 − θ(pF − p))|Φ〉

= gs

d3p

(2π)3e−ip(x−y)θ(x0 − y0)θ(p− pF ) , (8.7)

where we in the line have summed over λ giving rise to a degeneracy factor, which is denotedby gs. Using the integral reprensation for the step function (Exercise 5.6.4), we obtain

G(x, y) = gs

d3p

(2π)3

dp0

2πθ(p− pF )

e−iω(x0−y0)

p0 − ωp + iηeip·(x−y) . (8.8)

For y0 > x0, a similar calculation yields

G(x, y) = gs

d3p

(2π)3

dp0

2πθ(pF − p)

e−iω(x0−y0)

p0 − ωp − iηeip·(x−y) . (8.9)

The momentum space propagator then reads

G(p) = gs

[

θ(p− pF )

p0 − ωp + iη+

θ(pF − p)

p0 − ωp − iη

]

. (8.10)

Note that this reduces to the standard propagator in the vacuum , i .e. when pF = 0.

The density can now be written in terms of the propagator:

ρ = −i∫

d4p

(2π)4G(p)eiηp0

= −igs∫

d4p

(2π)4

[

θ(p− pF )

ω − ωp + iη+

θ(pF − p)

ω − ωp − iη

]

eiηp0 . (8.11)

Since η > 0, we must close the contour in the upper half plane such that the contributionfrom the semicircle vanishes as R → ∞. We then pick up a contribution from the pole

1The Fermi momentum is defined by ǫF =√

m2 + p2F . For a nonrelativistic particle, this reduces to

ǫF = p2F /2m.

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CHAPTER 8. FERMI GASES 91

ω = ωp + iη with residue 2πi. The density then reduces to

ρ = gs

d3p

(2π)3θ(pF − p)

=gs2π2

∫ ∞

0θ(pF − p)p2 dp

=gs6π2

p3F . (8.12)

This result is well known from e.g. TFY4230.

Using the same techniques as we did to express the density 〈ψ†ψ〉, we can express theexpectation value of the kinetic energy density T ≡ 〈ψ† 1

2m∇2ψ〉 in terms of the derivative

of the propagator. It is not so difficult to show that

T = ± limx′→x+

− 1

2m∇2G(x, t,x′, t+) , (8.13)

where + means taking the limit from above and t+ is infinitesimally larger than t 2. The ±sign apply to bosons or fermions, respectively. The latter ensures that pick up the contribu-tion from the pole in the upper half plane. For fermions, this yields

T = −igs∫ d4p

(2π)4

p2

2m

[

θ(p− pF )

ω − ωp + iη+

θ(pF − p)

ω − ωp − iη

]

eiηp0 . (8.14)

Closing the contour in the upper half plane and picking up the contribution from the poleω = ωp + iη, we find

T = gs

d3p

(2π)3

p2

2mθ(pF − p)

=gs2π2

∫ ∞

0θ(pF − p)

p4

2mdp

=gs

20π2mp5F . (8.15)

The energy per particle is given by T /ρ and so we obtain

E0 =p2F

2m

3

5. (8.16)

Thus the average energy of a particle is 3/5 times the Fermi energy.

Since we have not included interactions, T equals the total energy density E 3. The freeenergy density is

F = E − µρ

=gs

20π2mp5F − p2

F

2m

gs6π2

p3F

= − gs30π2m

p5F , (8.17)

2This ensures the correct ordering, i.e. ψ†ψ and not ψψ†.3Note that we are using ǫ = p2/2m and have subtracted the rest mass energy m. Taking it into account

yields an additional term in the energy density, mρ.

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CHAPTER 8. FERMI GASES 92

where we have used that µ = p2F/2m. The pressure P is minus the free energy density and is

a function of ρ. Inverting Eq. (8.12) to express pF in terms of ρ and inserting the expressioninto P = −F , we obtain

P =1

5m

(

6π2

gs

)2/3

ρ5/3 . (8.18)

which is the well known equation of state for an ideal Fermi gas at T = 0.

8.2 Interacting Fermi gas

The Lagrangian of a Fermi gas with a local two-body interaction with strength g is given by

Lint = iψ†∂0ψ − 1

2m(∇ψ†)(∇ψ) + µψ†ψ − 1

2g(ψ†ψ)2 . (8.19)

The perturbation Hint is given

Hint =1

2g∫

d3x (ψ†ψ)2 . (8.20)

The first-order correction to the ground-state energy, E1, is given by the expectation of theperturbation in the unperturbed ground state |Φ〉:

E1 = 〈Φ|Hint|Φ〉 . (8.21)

We can write

1

2g(ψ†ψ)2 =

g

2V 2

p,q,r,s,λ1,λ2

ei(p−q+r−s)xa†p,λ1aq,λ1

a†r,λ2as,λ2

, (8.22)

where we have used the orthogonality of χ(λi).

Integrating over x gives momentum conservation:

p− q + r − s = 0 . (8.23)

This yields

Hint =g

2V

q,r,s,λ1,λ2

a†q−r+s,λ1aq,λ1

a†r,λ2as,λ2

(8.24)

We first redefine the momentum, q′ = q − r, then s → p, drop the prime on the dummyvariable q′, and obtain

Hint =g

2V

p,q,r,λ1,λ2

a†p+q,λ1aq+r,λ1

a†r,λ2ap,λ2

(8.25)

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CHAPTER 8. FERMI GASES 93

We next separate the sum (8.25) into two terms, one with q = 0 and one with q 6= 0. Thefirst term reads

Haint =

g

2V

p,r,λ1,λ2

a†p,λ1ar,λ1

a†r,λ2ap,λ2

. (8.26)

Firstly, we note that we must have p < pF to get a nonzero result. We next distinguishbetween the cases r0 < pF and r > pF . In the first case, this implies that p = r. If wedenote the corresponding operator by Ha1

int, we obtain

Ha1int =

g

2V

p,λ1λ2

a†p,λ1ap,λ1

a†p,λ2ap,λ2

. (8.27)

The matrix element of this operator is

〈Φ|Ha1int|Φ〉 =

g

2V

p,λ1λ2

θ(pF − p)

→ 1

2gg2

s

d3p

(2π3)θ(pF − p) .

When we calculate the energy per particle, we first divide by the volume V to get the energydensity and then by ρ = p3

F/6π2. Hence this terms vanishes in the thermodynamic limit.

When r > pF , we must have λ1 = λ2 since ar,λ1must annihilate the particle above the

Fermi sea created by a†r,λ2. If we denote the corresponding operator by Ha2

int, the matrixelement becomes

〈Φ|Ha2int|Φ〉 =

g

2V

p,r,λ1

[1 − θ(pF − r)] θ(pF − p)

→ 1

2ggsV

d3p

(2π)3

d3r

(2π))3[1 − θ(pF − r)] θ(pF − p)

= −1

2ggsθ(pF − p)V

∫ d3r

(2π)3− 1

2ggsV

(

p3F

6π2

)2

. (8.28)

The term involving is divergent and is related to normal ordering and renormalization of thechemical potential. The second term is finite and gives the following total contribution tothe energy per particle from Ha

int:

Ea1 = −1

2g

(

p3F

6π2

)

. (8.29)

The second reads

Hbint =

g

2V

′∑

p,q,r,λ1,λ2

a†p+q,λ1aq+r,λ1

a†r,λ2ap,λ2

, (8.30)

where the prime on the sum is a reminder that the term q = 0 is omitted from the sum.Again we must have p < pF to get a nonzero contribution. If r > pF , we create a particle

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CHAPTER 8. FERMI GASES 94

above the Fermi sea and therefore this particle must be annihilated by the operator aq+r,λ1

(why?). Hence q = 0 but this is excluded. This implies that r < pF and therefore p = r

(why?). first. This yields

Hbint =

g

2V

′∑

p,q,λ1,λ2

a†p+q,λ1ap+q,λ1

a†p,λ2ap,λ2

, (8.31)

The energy them becomes

E1b =g

2Vg2s

′∑

p,q

θ(pF − p)θ(pF − |p + q|) . (8.32)

Taking the continuum limit, we obtain

E1b =1

2gV g2

s

d3p

(2π)3

d3q

(2π)3θ(pF − p)θ(pF − |p + q|) . (8.33)

Note that we have removed the prime from the integral as it only affects the integrand at asingle point. If we use dimensional regularization, we can redefine the momenta, q′ = q + p.The integrals then decouple and we obtain

E1b =1

2gV g2

s

(

p3F

6π2

)2

. (8.34)

The contribution to the energy per particle is then

E b1 =1

2ggs

p3F

6π2. (8.35)

Adding Eqs. (8.29) and (8.35), we obtain the total energy per particle to first order in theinteraction

E1 =p2F

2m

[

3

5+

2

3π(gs − 1)pFa+ ...

]

, (8.36)

where we have reinstated a = mg/4π. We notice that the expansion parameter is thedimensionless quantity pFa. Note also the a nonzero spin degeneracy is necessary to obtaina nonzero correction. The reason is the Pauli principle: The s-wave scattering length is zerofor two identical fermions, unless they have different spin.

8.3 Degenerate electron gas

So far, we have been discussing a degenerate Fermi gas with a local quartic interactioncharacterized by the s-wave scattering length a. We will next consider a nonrelativisticdegenerate electron gas in a volume V . The interaction is therefore the usual static Coulombinteraction between charges. In order to ensure that the system is overall electrically neutral,we introduce a homogeneous background of positive charge. The interactions we will takeinto account are thus

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CHAPTER 8. FERMI GASES 95

• Self-interaction of the homogeneous background.

• Electron-background interaction.

• Static electron-electron interaction.

From courses on electromagnetism, we know the expression for the Coulomb potential.However, we now have developed the machinery to derive it from QED. This is very instruc-tive and so we consider this first.

In Fig. 8.1, we have shown the e2-order Feynman diagram that contributes to electron-electron scattering. The photon which is exchanged between the two electrons is virtual andin order to evaluate the expression given by the Feynman graph, we need to know the photonpropagator.

Figure 8.1: Electron-electron scattering

8.3.1 Photon propagator and the Coulomb potential

The Lagrangian for free Maxwell theory is given by

L = −1

4FµνF

µν . (8.37)

The Green’s function for the photon field is a 4× 4 matrix and denoted by Dνβ(x− y). Theequation for the Green’s function reads

[∂α∂αgµν − ∂µ∂ν ]Dνβ(x− y) = iδµβδ

4(x− y) . (8.38)

In momentum space, this becomes[

−k2gµν + kµkν]

Dνβ(p) = iδµβ . (8.39)

This equation has no solution since the operator −k2gµν+kµkν is singular, i. e. det(−k2gµν+kµkν) = 0. The problem is that we have too many degrees of freedom. Recall that QED isa gauge theory which is invariant under local phase transformations of the fermion field ψaccompanied by a transformation of the gauge field Aµ. Hence there are (infinitely) many

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CHAPTER 8. FERMI GASES 96

gauge-field configurations that correspond to the same physical fields E and B 4. We musttherefore choose a gauge, i.e. somehow pick out one representative from each class of gauge-equivalent configurations. This is done by adding to the Lagrangian a gauge-fixing term 5

Lgf = − 1

2α(∂µA

µ)2 , (8.40)

where α is a so-called gauge parameter. The equation for the momentum-space propagatornow becomes 6

[

−k2gµν +(

1 − 1

α

)

kµkν]

Dνβ(p) = iδµβ . (8.41)

The solution is

Dµν(q) =igµνq2 + iǫ

− (1 − α)qµqν/q2

q2 + iǫ. (8.42)

Now it is important to stress that all physical quantities that one computes are independentof the gauge parameter α, and more generally independent of the gauge-fixing term in theLagrgangian 7. Two common and convenient choices are α = 1, which is called the Feynmangauge and α = 0, which is the Landau gauge.

We now return to the Feynman diagram Fig. 8.1. Up to some prefactors, the expressionfor it reads

Vijkl ∼ e2u(i)(p+ q)γµu(j)(p)

[

gµνq2

− (1 − α)qµqν/q2

q2 + iǫ

]

u(k)(k − q)γνu(l)(k) , (8.43)

where i, j, k, and l = 1, 2, and where the γ-matrices are are expressed in terms of the Paulispin matrices σi

γ0 =

(

1 00 −1

)

, γi =

(

0 σi−σi 0

)

, (8.44)

The spinors are given by

u(1)(p) =

10pz

E+mpx+ipy

E+m

, u(2)(p) =

01

px−ipy

E+m−pz

E+m

, (8.45)

4In fact, the space of gauge-field configurations can be divided into disjoint sets (equiavalence classes)that can be connected by a gauge transformation.

5This class of gauges is called covariant gauge. There exists another widely used class, the Coulombgauge where Lgf = − 1

2α(∂iA

i)2.6Exercise!7This is highly nontrivial at this stage.

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CHAPTER 8. FERMI GASES 97

We are interested in the nonrelativistic limit of the expression (8.43). In this limit, thespinors become

u(1)(p) =

1000

, u(2)(p) =

0100

, (8.46)

The indices i, j, k, and l of Vijkl then denote the spins of the electrons.

Using the expressions for the γ-matrices, it is easy to show that

u(i)(p)γ0u(j)(q) = δij u(i)(p)γku(j)(q) = 0 (8.47)

for all values of p and q and k = 1, 2, 3. This yields

Vijkl = e2δijg00

q2 + iǫ

[

1 − (1 − α)q20/q

2]

δkl . (8.48)

In the nonrelativistic limit, we have p ≈ (m,p) and (p + q) ≈ (m,p + q). This yieldsq ≈ (0,q) and therefore

Vijkl = e2δij1

q2δkl . (8.49)

Note that the α-dependence drops out in this limit The Fourier transform of this expression isessentially the Coulomb potential. Note that the Coulomb potential is time independent, i.e.it is static. It is an instantaneous interaction, which is reasonable since the nonrelativisticlimit is obtained by taking the limit c → ∞ and so it takes zero time for any signal orinteraction to propagate.

The Hamiltonian for the electron-electron interaction can be written as

Hel−el =1

2e2∫

d3x∫

d3y ψ†λ1

(x)ψλ1(x)

1

|x − y|ψ†λ2

(x)ψλ2(x) , (8.50)

where we have replaced the indices i, j, k, and l by the indices spin indices λ1 and λ2 andthe field ψ†

λ1(x) creates an electron with spin λ1 at position x.

8.3.2 Self-interaction of homogeneous background

The self-intercation energy of the homogeneous background of two infinitesimal regions inspace located at x and y with volumes d3x and d3y is

dVbg =1

2e2ρb(x)ρb(y)

|x − y| d3xd3y , (8.51)

where ρb(x) is the number density of the background at position x. The total interactionenergy Vb is found by integrating over x and y. Since the density of the background isassumed homogeneous, we obtain

Vbg =1

2e2ρ2

b

∫ d3xd3y

|x − y| . (8.52)

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CHAPTER 8. FERMI GASES 98

Changing variables x′ = x − y, we can write this as

Vbg =1

2e2ρ2

b

d3xd3y

|x|

=1

2e2ρ2

bV∫

d3x

|x| , (8.53)

where V is the volume of the system. The integral turn out to be divergent for large valuesof x and so we need to regulate it. We do this by adding a small mass µ to the photon 8.The propagator in momentum space then changes to that of a massive particle

1

q2→ 1

q2 + µ2. (8.54)

The Coulomb potential then changes to a Yukawa potential

1

|x| →e−µ|x|

|x| . (8.55)

The final integral in Eq. (8.53) is now easy to calculate and we finally obtain

Vbg =1

2e2ρ2

bV4π

µ2. (8.56)

8.3.3 Eletron interaction with homogeneous background

The interaction between the homogeneous background and a single electron located at y is

E1 = −e2∫

d3xρb

|x − y|e−µ|x−y| , (8.57)

where we have replaced the Coulomb potential with the Yukawa potential. After changingvariables, this yields

E1 = −e2ρb∫

d3xe−µ|x|

|x|

= −e2ρb4π

µ2. (8.58)

Note that this result is independent of the coordinate y of the electron (why?). The totalresult is then obtained by multlplying by N for the electron-background energy is then

Vb−el = −e2ρ2bV

µ2. (8.59)

8It is important to realize that regulating this divergence has nothing to do with the regularization ofdivergent integrals arising from Feynman graphs. We shall see that the divergence will cancel against anotherdivergent terms arising in the calculations and so regulating the integrals is only used in the intermediatesteps. The final result has no µ-dependence.

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CHAPTER 8. FERMI GASES 99

8.3.4 Eletron-electron interaction

We finallly consider the eletron-electron interaction term. Inserting the expression for thefield operators, this yields

Hel−el =e2

2V 2

d3x∫

d3y∑

p,q,r,s,λ1,λ2

a†p,λ1aq,λ1

1

|x − y|a†r,λ2

as,λ2ei(p−q)xei(r−s)y . (8.60)

We next need to evaluate

d3x∫

d3y ei(p−q)x1

|x − y|ei(r−s)y →

d3x∫

d3y ei(p−q)xe−µ|x−y|

|x − y| ei(r−s)y

(8.61)

=∫

d3x∫

d3ye−µ|y|

|y| ei(r−s)yei(p−q+r−s)x

=4π

(r − s)2 + µ2V δp−q+r−s , (8.62)

where we in the second line have changed variables. Inserting this result into Eq. (8.60) andsumming over s, we obtain

Hel−el =e2

2V

p,q,r,λ1,λ2

a†p,λ1aq,λ1

(p − q)2 + µ2a†r,λ2

ap−q+r,λ2

=e2

2V

p,q,r,λ1,λ2

a†p+q,λ1aq,λ1

p2 + µ2a†r,λ2

ap+r,λ2(8.63)

where we have changed variables, p′ = p−q. We are now ready to calculate the expectationvalue

Vel−el = 〈Φ|Hel−el|Φ〉 (8.64)

We first consider the case p = 0. This term reads

Hp=0el−el =

e2

2V

µ2

q,r,λ1,λ2

a†q,λ1aq,λ1

a†r,λ2ar,λ2

=1

2e2V

µ2

λ1,λ2

d3q

(2π)3

d3r

(2π)3a†q,λ1

aq,λ1a†r,λ2

ar,λ2, (8.65)

where we in the last line have taken the large-volum limit. This operator is proportional toN2. This immediately gives

Ep=0el−el =

1

2Ve2N2 4π

µ2. (8.66)

The interpretation of this contribution is given by the Feynman diagram to the right inFig. 8.2.

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CHAPTER 8. FERMI GASES 100

Figure 8.2: Order-e2 vacuum diagrams contributing to the electron-electron interaction en-ergy.

Adding this result to Eqs. (8.56) and (8.58), we notice that the terms that are individuallydivergent in the limit µ→ 0 cancel 9.

We next consider the p 6= 0 term in the sum

Hp6=0el−el =

e2

2V

′∑

p,q,r,λ1,λ2

a†p+q,λ1aq,λ1

p2 + µ2a†r,λ2

ap+r,λ2(8.67)

where the prime is a reminder that p = 0 is excluded. The contribution is given by theleft Feynman graph in Fig. 8.2. By the usual arguments, we must have |p + r| < pF and|r| > pF . This is implies that p + r = p + r, i. e. q = r. This yields

Vel−el =e2

2V

′∑

p,q,λ1

p2θ(pF − |p + q|) [1 − θ(pF − |q|)]

= 4πe2V∫

d3p

(2π)3

d3q

(2π)3

1

p2θ(pF − |p + q|) [1 − θ(pF − |q|)] . (8.68)

Notice that the prime is removed from the integral as contributes to a single point only. Asusual, the first term is divergent and related to normal ordering. We discard it and are leftwith 10

Vel−el = −4πe2V∫

d3p

(2π)3

d3q

(2π)3

1

(p− q)2θ(pF − |p|)θ(pF − |q|) , (8.69)

where we have changed variables. We therefore need to calculate the angular average of1/|q − p|. This is

1

|q − p|

= 2π∫ π

0

sin θ

p2 + q2 + 2|p||q| cos θdθ

|p||q| log(|p| + |q|)2

(|p| − |q|)2, (8.70)

The radial integrals then become

1

16π4

∫ pF

0dp

∫ qF

0dq pq log

(|p| + |q|)2

(|p| − |q|)2=

1

16π4

∫ pF

0dp

[

pFp2 +

1

4p(p2

F − p2) log(pF + p)2

(pF − p)2

]

=1

16π4p4F . (8.71)

9The terms with p 6= 0 from Hel−el are clearly convergent in the limit µ→ 0.10Note that the way we calculate the remaining integrals is very different from the arguments given in

Fetter and Walecka.

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CHAPTER 8. FERMI GASES 101

This yields

Vel−el = −e2V

4π3p4F . (8.72)

This is the final result for the interaction energy of a degenerate electron plasma and is inagreement with the result of Fetter and Walecka, Eq. (3.36).

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Chapter 9

Effective potential

In this section, we introduce the concept of the effective potential in the context of theharmonic oscillator. We then apply the formalism to the weakly interacting Bose. In bothcases, we rederive well-known results.

9.1 Harmonic oscillator

Consider a harmonic oscillator in one dimension with the Lagrangian

L =1

2mx2 − 1

2mω2x2 . (9.1)

Scaling by setting φ =√mx, we can write

L =1

2φ2 − 1

2ω2φ2 . (9.2)

Recall that quantum mechanics is field theory in 0 + 1 dimensions and we therefore expandthe field φ as φ0 + φ. This yields

L =1

2˙φ2 − 1

2ω2(φ2

0 + 2φ0φ+ φ2) . (9.3)

The classical potential is defined by

V0 =1

2ω2φ2

0 , (9.4)

and is simply the (scaled) potential energy of the oscillator as a function of φ0. The propa-gator in momentum space can be computed using standard techiniques and reads 1

D(p0) =i

p20 − ω2 + iǫ

. (9.5)

1D does not depend on any spatial momenta since we are in 0+1 dimensions.

102

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CHAPTER 9. EFFECTIVE POTENTIAL 103

Note that the linear term in L vanishes at the minimum of the classical potential (as usual)i.e. for φ0 = 0 and the classical energy of the oscillator vanishes.

One can systematically calculate corrections to the classical potential The potential isthen called the effective potential and is the classical potential plus quantum corrections. Ifthe theory is a free field theory, one can calculate the quantum corrections exactly, whichwe will now do for the harmonic oscillator. The first quantum correction V1 to the classicalpotential is a so-called one-loop correction and reads

V1 = − i

2

dp0

ddp

(2π)dln[−i detD−1(p0, p)] , (9.6)

where d is the number of spatial dimensions and D−1(p0, p) is the inverse propagator inmomentum space. For a free field theory

V = V0 + V1 (9.7)

is exact. For the oscillator, d = 0 and we obtain

V1 = − i

2

dp0

2πln[ω2 − p2

0 − iǫ] .

In order to calculate V1, we first consider the derivative of V1 with respect to ω2:

I =i

2

dp0

1

(p20 − ω2 + iǫ)

=1

4

1

ω, (9.8)

where we have used standard contour integration 2. Integrating with respect to ω2 we obtain 3

V1 =1

2ω . (9.9)

Thus the complete effective potential is

V =1

2ω2φ2

0 +1

2ω . (9.10)

The ground-state energy E0 is given by the minimum of the quantum effective potential.The value of φ0 which minimizes V is φ0 = 0 and so

E0 = V (φ0 = 0)

=1

2ω . (9.11)

2Close the contour in either halfplane. The contribution comes from one of the poles p0 = ±(ω− iǫ) andthe contribution from the semicircle vanishes.

3We set the integration constant to zero!

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CHAPTER 9. EFFECTIVE POTENTIAL 104

Reinstating factors of h one finds

E0 =1

2hω . (9.12)

Thus the classical energy of a harmonic oscillator is zero (the oscillator is at rest at theorigin, φ0 = 0), while the one-loop correction includes the zero-point fluctuations and givesthe exact quantum mechanical ground-state energy.

For interacting field theories, one can write the effective potential in a perturbative seriesas

V = V0 + V1 + V2 + ... (9.13)

V2, V3... are corrections to the exactly solvable case with no interactions, i.e. a theory thatis quadratic in the fields.

9.2 Dilute Bose gas revisited

The Lagrangian of the dilute Bose gas is

L = iψ†∂0ψ + µψ†ψ − 1

2m∇ψ† · ∇ψ − 1

2g(

ψ†ψ)2

, (9.14)

Writing the ψ as a sum of a classical field and a quantum fluctuating field

ψ = φ0 + ψ , (9.15)

one finds

L = iψ†∂0ψ − 1

2m∇ψ† · ∇ψ + µ

[

|φ0|2 + φ∗0ψ + φ0ψ

† + ψ†ψ]

−1

2g[|φ0|2 + φ∗

0ψ + φ0ψ† + ψ†ψ]2 , (9.16)

where we have neglected total derivatives and used that φ0 is independent of space-time.The complex field ψ is written in terms of two real fields

ψ =1√2(ψ1 + iψ2) . (9.17)

Inserting this into the Lagrangian, we obtain

L =1

2(ψ2∂0ψ1 − ψ1∂0ψ2) +

1

2

1

2m(ψ1∇2ψ1 + ψ2∇2ψ2) +

[

µ− 1

2g|φ2

0|]

|φ0|2

+1√2

[

µ− 1

2g|φ0|2

]

φ∗0[ψ1 + iψ2] +

1√2

[

µ− 1

2g|φ0|2

]

φ0[ψ1 − iψ2]

+1

2(µ− 3g|φ0|2)ψ1

2+

1

2(µ− g|φ0|2)ψ2

2, (9.18)

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CHAPTER 9. EFFECTIVE POTENTIAL 105

where we have integrated by parts and neglected terms that are third and fourth in thefluctuations. The classical potential is

V0 =[

µ− 1

2g|φ2

0|]

|φ0|2 . (9.19)

Note that the linear terms vanish at the minimum of the classical potential, µ = g|φ0|2, andthat V0 reduces to

V0 = µ2/2g . (9.20)

The terms that are quadratic in the fluctuations can be written in matrix form

Lquad = −1

2(ψ1, ψ2)

(

− 12m

∇2 − µ+ 3g|φ0|2 −∂0

∂0 − 12m

∇2 − µ+ g|φ0|2)(

ψ1

ψ2

)

.(9.21)

The matrix is denoted by G−1 and defines the inverse propagator. In momentum space, G−1

reads

iG−1(p, ω) =

(

p2

2m− µ+ 3g|φ0|2 −iω

iω p2

2m− µ+ g|φ0|2

)

. (9.22)

The determinant is

det iG−1 =

[

p2

2m− µ+ g|φ0|2

] [

p2

2m− µ+ 3g|φ0|2

]

− ω2 . (9.23)

The zeros of det iG−1 yields the dispersion relation. This gives

ω2 =

[

p2

2m− µ+ g|φ0|2

] [

p2

2m− µ+ 3g|φ0|2

]

. (9.24)

Evaluated at the classical minimum µ = g|φ0|2, we find

ǫ2(p) =p2

2m

[

p2

2m+ 2g|φ0|2

]

, (9.25)

which is the Bogoliubov dispersion relation derived earlier.

The one-loop effective potential is given by

V1 = − i

2

dp0

ddp

(2π)dln det[−iG−1(p0, p)]

=∫

dp0

ddp

(2π)dln[ǫ2(p) − p2

0] . (9.26)

Integrating over p0 yields 4

V1 =1

2

ddp

(2π)dǫ(p) . (9.27)

4Take the derivative of V1 wrt ǫ2(p) and do the contour integral of dV1/dǫ2(p) in the complex p0 plane.

Integrate the resulting expression wrt ǫ2(p) and set the constant of integration to zero.

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CHAPTER 9. EFFECTIVE POTENTIAL 106

The expression for V1 is highly divergent in the UV. In fact, for d = 3 it contains quinticcubic, and linear divergences. This is seen by expanding the integrand for large p. Afterintegrating over angles, we obtain

V1 =1

4π2

∫ ∞

0dp p2ǫ(p) (9.28)

For large-p, the integrand behaves as

p2ǫ(p) ≃ p2

[

p2

2m+ µ− µ2

p2/m+ ...

]

(9.29)

We isolate the divergences by adding and subtracting the infinite terms

V1 =1

4π2

∫ ∞

0dp p2

[

ǫ(p) − p2

2m− µ+

µ2

p2/m

]

+1

4π2

∫ ∞

0dp p2

[

p2

2m+ µ− µ2

p2/m

]

. (9.30)

The first divergence corresponds the infinite vacuum energy density of an ideal gas and reads

Evacuum =1

2

d3p

(2π)3

p2

2m. (9.31)

The counterterm that arises from the classical potential is

∆V0 = ∆E − µ

g∆µ+

µ2

2g2∆g , (9.32)

where we have added a vacuum counterterm ∆E . This term in Eq. (9.31) is eliminated bythe vacuum counterterm which becomes

∆E = −1

2

d3p

(2π)3

p2

2m. (9.33)

This divergence is related to the operator-ordering problem in the formalism 5. The secondterm is

1

2µ∫

d3p

(2π)3. (9.34)

This term is cancelled by the counterterm

−µg

∆µ , (9.35)

5Classically the Fourier coefficients a and a∗ commute and so aa∗ is the same a∗a. This is no longer thecase in the quantum theory where aa† 6= a†a!

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CHAPTER 9. EFFECTIVE POTENTIAL 107

and so matching the two terms yields

∆µ =1

2g∫

d3p

(2π)3. (9.36)

The last divergence is

−1

2mµ2

d3p

(2π)3

1

p2. (9.37)

This divergence is cancelled by the counterterm

µ2

g2∆g . (9.38)

Mathcing the two expressions, we find

∆g = mg2∫

d3p

(2π)3

1

p2. (9.39)

Making the subtractions, the renormalized free energy density is

V1 =1

2π2

∫ ∞

0dp p2

[

ǫ(p) − p2

2m− µ+

µ2

p2/2m

]

=2√

2

15π(2m)3/2µ5/2 . (9.40)

The free energy density is the sum of the classical potential V0 = −µ|φ0|2 + g|φ0|4/2 and V1

evaluated at the minimum µ = g|φ0|2 and expressed as a function of µ. This yields

F = −µ2

2g

[

1 − 4(2m)3/2√

2µg2

15π2

]

. (9.41)

This result is identical to that obtained by first calculating the energy density E and thenperforming a Legendre transform to obtain F .

9.3 Problems

9.3.1

Consider a relativistic scalar field theory with Lagrangian

L =1

2(∂µφ)(∂µφ) − 1

2m2φ2 − λ

4φ4 . (9.42)

1) Write the field as φ = φ0 + φ and show that the classical potential is

V0 =1

2m2φ2

0 +λ

4φ4

0 . (9.43)

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CHAPTER 9. EFFECTIVE POTENTIAL 108

2) The propagator is

D =i

p2 −M2, (9.44)

where M2 = m2 + 3λφ20. Show that the one-loop correction to the effective potential is

V1 =1

2

ddp

(2π)d

|p|2 +M2 , (9.45)

where d is the number of spatial dimensions.

3) Set d = 2. The expression for V1 is divergent. Analyze the divergences and subtractcounterterms such that the renormalized effective potential is finite. Hint: you only need torenormalize the vacuum energy density and the mass parameter. The coupling λ needs norenormalization. In other words, write

m2 → m2 + ∆m2 , (9.46)

and calculate ∆m2.

4) Set d = 3 and repeat using dimensional regularization.

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Chapter 10

Bose-Einstein condensation in traps

Figure 10.1: Images of the velocity distribution of rubidium atoms in the experiment byAnderson et al. (1995). The left frame corresponds to a gas at a temperature just abovecondensation; the center frame, just after the appearance of the condensate; the right frame,after further evaporation leaves a sample of nearly pure condensate.

Bose-Einstein condensation in harmonic traps is (simply) a many-body problem in anexternal potential V (x). The potential we will be using in the following is a three-dimensionalharmonic potential

V =1

2m(ω2

xx2 + ω2

yy2 + ω2

zz2) . (10.1)

In most cases, we restrict ourselves to a spherically symmetric potential, i.e. ωx = ωy = ωz.

The starting point is the action

S =∫

dtd3x[

iψ†(x, t)∂tψ(x, t) + (µ− V (x))ψ†(x, t)ψ(x, t) − 1

2m∇ψ†(x, t) · ∇ψ(x, t)

−1

2g(

ψ†(x, t)ψ(x, t))2]

. (10.2)

109

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 110

The new term V (x)ψ†ψ comes about since ψ†ψ is the density operator and V (x) is thepotential energy of a particle at the position x. Hence V (x)ψ†ψ is the potential energydensity due to the trapping potential.

Mean-field theory is obtained by replacing ψ = φ0 + ψ by φ0 and ignoring quantumfluctuations 1. The classical equation of motion for the field φ0 is

i∂φ0

∂t= − 1

2m∇2φ0 − [µ− V (x)]φ0 + gρφ0 , (10.3)

where have used that ρ = φ20 in the mean-field approximation. Note that [µ − V (x)] acts

as a local space-dependent effective chemical potential. This equation is called the time-

dependent Gross-Pitaevskii equation.

The ground state of the trap is independent of time and satifies Eq. (10.3) with a vanishingleft-hand-side:

0 = − 1

2m∇2φ0 − [µ− V (x)]φ0 + gρφ0 . (10.4)

From this equation, it is clear that the ground state is no longer homogeneous in space. Infact, turning off the interactions, we know that the single-particle ground state is a Gaussian(the ground state of a single oscillator). In the many-particle ground state all particlesreside in the single-particle ground state and the many-body wavefunction is a product ofN Gaussians.

Assume that the spatial extent of the cloud is r0. In the absence of interactions, the kineticterm gives a contribution to the energy per particle ∼ 1/2mr2

0 since a typical momentumis p ∼ 1/r0 which follows from the uncertaintly principle. Similarly, the potential termgives a contribution to the energy per particle on the order mω2

0r20/2. Balancing these

contributions, i.e. minimizing the total energy, gives r0 ∼ 1/√mω0, which is the length scale

l of the harmonic oscillator. Now consider interactions. If there are N particles in the trap,the density is ρ ∼ N/r3

0 and the interaction energy per particle is on the order gρ ∼ gN/r30.

The interaction term thus shifts the radius of the cloud towards larger values 2.

If we assume that interactions are dominant 3, we can ignore the differential operator∇2/2m in Eq. (10.4). This approximation is called the Thomas-Fermi approximation. Wecan then find an explicit expression for the density by solving Eq. (10.4) ignoring the firstterm. For a spherically symmetry trapping potential V = mω2

0r2/2, this yields

ρ =

1g[µ− V ] , r < r0

0 , r > r0, (10.5)

where r0 satisfies

1

2mω2

0r20 = µ . (10.6)

1In a trap φ0 does depend on x.2This is correct if g > 0, i.e. if interactions are repulsive. We assume that g > in the following.3Thus we are considering the strong-coupling limit which is obtained by letting gN → ∞.

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 111

The solution is thus an upside-down parabola. Given the trapping potential V (x), one candetermine the chemical potential µ by using the normalization condition

d3xρ = N , (10.7)

where N is the total number of particles in the trap. Integrating over angles, this yields

∫ r0

0[µ− 1

2mω2

0r2]r2dr =

gN

4π. (10.8)

Integrating this expression, we obtain

gN

4π=

1

3µr3

0 −1

10mω2

0r50

=1

15mω2

0r50 , (10.9)

where we have substituted µ = mω2r20/2. Solving with respect to r0, we find

r0 =(

15Na

l

)1/5

l , (10.10)

where l = 1/√mω0 is the length scale of the oscillator. The chemical potential then becomes

µ =1

2mω2

0r20

=1

2

(

15Na

l

)2/5

ω0 . (10.11)

In Fig. 10.2, we show the density distribution of 80000 sodium atoms in the trap of Hauet al. (1998) as a function of the axial co-ordinate. The experimental points correspondto the measured optical density, which is proportional to the column density of the atomcloud along the path of the light beam. The dashed line is the noninteracting ground state,i.e. a Gaussian profile. The solid line is the up-side down parabola, i.e. the Thomas-Fermiapproximation. The agreement with the data points is impressive The figure points out therole of atom-atom interaction in reducing the central density and enlarging the size of thecloud.

10.1 Hydrodynamics of BEC

In this section, we briefly discuss the excitations around the ground state for a trapped Bosegas.

The equation of motion for the quantum field ψ is

iψ =[

− 1

2m∇2 + V − µ

]

ψ + g(ψ†ψ)ψ . (10.12)

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 112

Figure 10.2: Density profile in the z-direction. See main text for details.

The quantum field is written as the sum of a classical time-dependent field φ0 and a quantumfluctuating field ψ as

ψ = φ0 + ψ . (10.13)

The expectation values of density ρ and current density j are

ρ = 〈ψ†ψ〉= |φ0|2 + 〈ψ†ψ〉 , (10.14)

j = − i

2m〈ψ†∇ψ − ψ∇ψ†〉

= − i

2m(φ∗

0∇φ0 − φ0∇φ∗0) −

i

2m〈ψ†∇ψ − ψ∇ψ†〉 . (10.15)

Taking the expectation value of the equation of motion (10.12) , we obtain

iφ0 =[

− 1

2m∇2 + V − µ

]

φ0 + g[

|φ0|2φ0 + 2〈ψ†ψ〉φ0 + 〈ψψ〉φ∗0 + 〈ψ†ψψ〉

]

,(10.16)

where we have used that 〈ψ†〉 = 〈ψ〉 = 0. In the mean-field approximation, the matrixelements are neglected and the equation reduces to the time-dependent Gross-Pitaevskiiequation:

iφ0 =[

− 1

2m∇2 + V − µ

]

φ0 + g|φ0|2φ0 . (10.17)

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 113

Multiplying Eq. (10.17) by φ∗0 and adding its complex conjugate i.e. adding φ∗

0φ0, we obtain

i∂|φ0|2∂t

+ ∇ ·[

1

2m(φ∗

0∇φ0 − φ0∇φ∗0)]

= 0 .

Since ρ = |φ0|2 in the mean-field approximation, Eq. (10.18) is simply the continuity equationρ + ∇ · j, where the current density j is given by its mean-field approximation, i.e. byEq. (10.15) neglecting the matrix elements. If we define the velocity v by j/ρ, we can write

∂ρ

∂t+ ∇ · (ρv) = 0 . (10.18)

We next write the field φ0 in terms of its magnitude√ρ and its phase φ as φ0 =

√ρeiφ 4.

Inserting this into the expression for j, we obtain

j = − i

2m

[

ρe−iφ∇eiφ − ρeiφ∇e−iφ]

m∇φ . (10.19)

Comparing this with j = ρv, we find

v =1

m∇φ . (10.20)

Inserting φ0 =√ρeiφ into the equation of motion and taking the real part of the resulting

expression, one finds

−∂φ∂t

= − 1

2m√ρ∇2√ρ+

1

2mv2 + gρ+ V − µ . (10.21)

Taking the gradient of this equation and the time derivative of ∇φ = mv and equating theresult we find an expression for m∂v/∂t.

m∂v

∂t= −∇

(

µ+1

2mv2

)

, (10.22)

where

µ = V + gρ− µ− 1

2m√ρ∇2√ρ . (10.23)

Eqs. (10.18) and (10.22) constitute the hydrodynamic equations for the problem since theyinvolve the hydrodynamic variables density ρ and velocity v.

4This is called the polar representation of the field for obvious reasons. This is in contrast to the cartesianapproximation used earlier.

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 114

We are now going to study oscillations on top of the ground state in the Thomas-Fermilimit. We can then neglect the last term in Eq. (10.23) to a first approximation. Taking thetime derivative of the continuity equation, we obtain

ρ+ ∇ · (ρv + ρv) ≈ ρ+ ∇ · (ρv)

= 0 , (10.24)

where we neglect the term ρv since it is higher order in derivatives. Inserting the expressionfor v into Eq. (10.24), we find

ρ =1

m∇ · [ρ∇(V − µ+ gρ)] , (10.25)

where we also neglect the term mv2/2 treating v as a small quantity. The density is nowwritten as a sum of the ground-state density and a time-dependent fluctuation:

ρ = ρ0 + δρ . (10.26)

Inserting this into Eq. (10.25), we obtain to first order in the perturbation δρ

δρ =g

m∇ · [ρ0∇δρ] , (10.27)

where we have used that V − µ = ρ0. Assuming the confining potential is sphericallysymmetric and the time dependence of δρ ∝ e−iωt, this can be written as

ω2δρ = ω20r∂δρ

∂r− 1

2ω2

0(r20 − r2)∇2δρ , (10.28)

where ω0 denotes the oscillator frequency and where we have used that ρ0 = [µ−V ]/g. Thespherical symmetry implies that the solutions can be written as

δρ = D(r)Ylm(θ, φ) . (10.29)

A simple class of solutions is

δρ = CrlYlm(θ, φ) , (10.30)

where C is a constant. The radial part of second term on the right-hand-side of Eq. (10.28)is

[

1

r2

∂rr2 ∂

∂r− l(l + 1)

r2

]

rl = 0 . (10.31)

The implies that

ω2 = ω20l . (10.32)

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CHAPTER 10. BOSE-EINSTEIN CONDENSATION IN TRAPS 115

The solution l = 0 corresponds to a constant change of ρ everywhere and therefore a constantchange in the chemical potential µ. The frequency is of course zero in this case. The solutionsl = 1 correspond to a translation of the cloud with no change in internal structure. Considerfor example l = 1 and m = 0, i.e.

δρ = CrY10

∝ z . (10.33)

The ground-state density ρ0 of the cloud goes like (1 − r2/r20) and so a small translation in

the z-direction is therefore proportional to δρ ∝ ∂ρ0∂z

∝ z. Higher-order solutions are morecomplicated.

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Appendix A

Sum vs integral

If we use a large quantization volume V = L3 and periodic boundary conditions, the solutionsto the KG-equations are

φk(x, t) ∼ e−ikx , (A.1)

where

kx =2π

Ln , (A.2)

where n = 0, 1, 2, ... with similar expresssions for ky and kz. The normalized solutions are

φk(x, t) =1√Veikx . (A.3)

The inner product is

I =∫

d3xφ∗k(x, t)φp(x, t)

=1

Vei(k0−p0)t

e−i(k−p)·x d3x . (A.4)

If p = k, clearly the integral gives a factor of V . If p 6= k, the integral vanishes due to theperiodic boundary conditions. This yields

I = ei(k0−p0)tδp,k

= δp,k , (A.5)

where we have used k0 = p0 whenever p = k. The integral is now being replaced by a sumusing the prescription

d3p

(2π)3→ 1

V

p

. (A.6)

116

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APPENDIX A. SUM VS INTEGRAL 117

Let us check that this is consistent. The equal-time commutator can be written as

[φ(x, t), π(y, t)] =1

V

p,q

(

− i

2

)

[

[ap, aq]e−i(px+qy) − [a†p, a

†q]e

i(px+qy)

+[a†p, aq]ei(px−qy) − [ap, a

†q]e

−i(px−qy)]

. (A.7)

We next assume the commutation relations are as for a single oscillator, i.e.[

ap, a†q

]

= δp,q (A.8)

[ap, aq] = 0 , (A.9)[

a†p, a†q

]

= 0 . (A.10)

This yields

[φ(x, t), π(y, t)] =i

V

p,q

EqEpδp,qe

−i(px−qy)

=i

V

p

eip·(x−y) . (A.11)

Eq. (A.11) is simply the representation of δ3(x − y) for discrete values of p. Let us checkthis. We define

f(x) =1

V

p

eip·x (A.12)

Integrating over x, yields

d3xf(x) =1

V

d3x∑

p

eip·x

=∑

p

δp,0eip·x

= 1 ,

where we again have used the periodic boundary conditions in the second line. The textbook of Fetter and Walecka is using discrete sums instead of integrals and this little notegives the recipe to switch.

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Appendix B

Interatomic potentials and scattering

processes

So far we have been considering noninteracting systems, with a few exceptions such as a Diracfermion in an external Coulomb potential. We are now going to discuss interacting systems.An important application is the dilute Bose gas and therefore we need to discuss interatomicpotentials. A general discussion on scattering can be found Ref. [9] and a discussion onatomic potentials and the dilute Bose gas can be found in Ref. [13].

The action is

S =∫

dt ∫

d3xψ†(x, t)[

i∂t +1

2m∇2 + µ

]

ψ(x, t)

−1

2

d3x∫

d3x′ ψ†(x, t)ψ†(x′, t)V0(x − x′)ψ(x, t)ψ(x′, t) + ...

. (B.1)

Here ψ†(x, t) is a quantum field that creates a boson at the position x and time t and ψ(x, t)is a quantum field that annihilates an atom at the position x and time t. µ is the chemicalpotential which can be used to tune the average particle density 1, and V0(x) is the two-bodypotential. Note that the two-body potential is static, i.e. independent of time. This implieswe treat the interactions as being instantaneous. Finally, the ellipsis indicate that we areneglecting three-body potentials etc.

The interatomic potential can be divided into a central part V c0 (x) and a remainder. The

central part of the potential depends only on the separation of the atoms and their electronicspins. It conserves separately the total orbital and the total electronic spin of the atoms.The noncentral part conserves the total angular momentum but does not separately conservethe total orbital or the total electronic spin of the atoms. An example of a noncentral partof the interaction is the dipole-dipole interaction:

Vdipole−dipole =1

4πǫ0x3[d1 · d2 − 3(d1 · x)(d2 · x)] , (B.2)

1Recall from statistical mechanics that the µ is introduced in the grand canonical ensemble where theparticle number is not fixed.

118

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 119

where di (i = 1, 2) are the dipole moments and x = x/|x|. The central part of the potentialconsists of a short-range part with range x0 and a long-range Van der Waals tail. The lattergoes as 1/x6 as x→ ∞ 2.

A typical two-body potential is shown in Fig. B.1.

r

V(r)

Figure B.1: Typical central part of an interatomic potential.

Now a real interatomic is very difficult to handle in practice. For example, in order to findthe bound state of a potential V0(x − x′), we would have to solve the Schrodinger equationthat follows from the action (B.8):

[

− 1

2m∇2 +

d3x′ψ†(x′, t)V0(x − x′)ψ(x′, t)]

ψ(x, t) = Eψ(x, t) , (B.3)

where E is the energy of the bound state and we have set µ = 0. This is an integral equationand is nonlocal 3 and therefore numerically very difficult to solve in practice. It is thereforecommon to simplify the problem by introducing some model potentials that mimics thephysics of a real potential. The model potentials typically have one or more parameters thatcan be adjusted to reproduce the physics of a real potential. We will return to this pointbelow.

2If retardation effects are taken into account, it goes as 1/x7.3Nonlocal means that the value of the ψ(x, t) depends on the values of not only in x but on other values

as well.

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 120

A model potential of this kind is the sum of a hard-core potential with range x0 and avan-der-Waals potential:

V c0 (x) =

+∞ , x < x0

−C6

x6 , x > x0

, (B.4)

where C6 is a constant. Another example is the hard-core square-well potential

V c0 (x) =

+∞ , x < xc−V0 xc < x < x0

−C6

x6 , x0 < x, (B.5)

where xc is a constant.

In Fig. B.2, we have shown the Fourier transform V (k) of a typical short-range potentialwith range x0. Note in particular that the Fourier transforms vanishes in the limit k → ∞.

V(k)

Mk

Figure B.2: Fourier transform V (k) interatomic potential.

The simplest model potential is a modification of the potential (B.5), where we neglectthe Van de Waals tail and set x0 = 0. The potential is then a contact potential in real spaceof strength g and can be written as

V0(x − x′) = gδ(x − x′) . (B.6)

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 121

The Fourier transform is

V0(q) =∫

d3xeiq·xV0(x − x′)

=∫

d3xgδ(x − x′)

= g , (B.7)

i.e. it is constant in momentum space. This is shown by the dotted line in Fig. (B.2). Thecontact potential is singular in coordinate space and is constant in momentum space. Thisleads to some technical problems with diverging integrals and the need to renormalize thetheory 4. This is related to the fact that V (k) does not vanish for k → ∞.

Inserting the potential (B.6) into the action, neglecting three-body potentials, and inte-grating over x′, we obtain

S =∫

dt ∫

d3xψ∗(x, t)[

i∂t +1

2m∇2 + µ

]

ψ(x, t) − 1

2g(

ψ†(x, t)ψ(x, t))2

.

This action will be the starting point for our treatment of the weakly interacting Bose gas.

B.1 Scattering and the Born approximation

In this section, we are very briefly going to discuss scattering of an atom off of a time-independent potential. This is equivalent to two-particle scattering processes where theparticles interact via a two-body potential V0(x). The reduced mass of the particle is m∗ =m/2. For a detailed account on scattering, consult Hemmer or any other standard textbookon quantum mechanics.

The scattering process is described by the time-independent Schrodinger equation:[

− 1

2m∗∇2 + V0(x)

]

ψ(x) = Eψ(x) , (B.8)

where E is the energy of the particle. The boundary conditions for this problem must besuch that it describes an incoming ray of particles and outgoing scattered particles. Forlarge |x|, the wavefunction is therefore a sum of an incoming plane wave and an outgoingspherical wave

ψ ∼ eik·x + f(θ, φ)eikx

x, (B.9)

where f(θ, φ) is the scattering amplitude.

We next introduce E = k2/2m∗ and U(x) = 2m∗V0(x). This yields[

∇2 + k2]

ψ(x) = U(x)ψ(x) . (B.10)

4We have already seen that the ground-state energy of a quantum field theory is divergent, but weneglected the divergence. In interacting field theories, more divergences appear and we tame them byredefining the coupling constant g, as we shall see.

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 122

The differential equation (B.10) is equivalent to the integral equation

ψ(x) = ψ0(x) +∫

d3x′G0(x − x′)U(x′)ψ(x′) , (B.11)

where ψ0(x) is a solution to the homogeneous equation (∇2 + k2)ψ0(x) = 0 and the free-particle Green’s functions G0(x − x′) satisfies

[

∇2 + k2]

G0(x − x′) = δ(x − x′) . (B.12)

The equivalence of these equations is easy to chech by acting on equation (B.10) with ∇2+k2,using that (∇2 + k2)ψ0(x) = 0 and that G0(x − x′) satisfies Eq. (B.12). The solution toEq. (B.12) is given by

G0(x − x′) = − eik|x−x′|

4π|x − x′| , (B.13)

and in momentum space the free propagator is

G0(p) =1

k2 − p2 + iη, (B.14)

Collecting the various pieces, the integral equation (B.11) can be written as

ψ(x) = eik·x − 1

d3x′eik|x−x′|

|x − x′|U(x′)ψ(x′) . (B.15)

For large values of x, we can expand the exponent and write

k|x − x′| ≈ kx− k′ · x′, (B.16)

where k′ ≡ kx/x. This implies that the asymptotic wavefunction can be written as

ψ(x) ≈ eik·x − 14π

eikx

x

d3x′e−ik′·x′

U(x′)ψ(x′) . (B.17)

The scattering amplitude is therefore

f(k′,k) = − 1

d3x′e−ik′·x′

U(x′)ψ(x′) . (B.18)

The Fourier transform of G0(x − x′), ψ(x), and U(x) are given implicitly by

G0(x − x′) =∫

d3p

(2π)3e−ip·(x

′−x) 1

k2 − p2 + iη, (B.19)

ψ(x) =∫

d3p

(2π)3eip·xψ(p), (B.20)

U(x) =∫ d3p

(2π)3eip·xU(p) , (B.21)

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 123

By going to momentum space, Eqs. (B.11) and (B.18) can be written as

ψ(p) = (2π)3δ(p− k) +1

k2 − p2 + iη

d3q

(2π)3U(q)ψ(p − q); ,

f(k′,k) = − 1

d3q

(2π)3U(q)ψ(k′ − q). (B.22)

By combining these equations, we can write

ψ(p) = (2π)3δ(p− k) − 4πf(p,k)

k2 − p2 + iη. (B.23)

Multiplying Eq. (B.23) by −U(k′ − p) and integrating with respect to p, we obtain thefollowing integral equation for the scattering amplitude

4πf(k′,k) = −U(k′ − k) + 4π∫

d3p

(2π)3

U(k′ − p)

k2 − p2 + iηf(p,k). (B.24)

This equation can be solved by iteration. The first approximation to f(k′,k) is found byignoring the second term on the right-hand-side of (B.24). This yields

f(k′,k) = − 1

4πU(k′ − k) . (B.25)

This approximation is called the first Born approximation.

If we take the limit k → 0, the partial wave method yields

f(0, 0) = −a , (B.26)

where a is the scattering length. The scattering length is an observable that one can measurein experiments. The idea of using an effective potential or a model potential is to tune theparameters in the potential such that one reproduces the correct observed scattering length.This procedure is called matching. Using the delta-function potential, which in momentumspace is U(k − k′) = 2m∗g, we find the in the Born approximation

f(k′,k) = −m∗g

2π, (B.27)

and therefore

a =m

4πg , (B.28)

where we have used m∗ = m/2. Using the first-order solution f(k′,k) = −mg/4π in theintegral equation and iterating, we find the second approximation

4πf(k′,k) = −mg −mg2∫ d3p

(2π)3

1

k2 − p2 + iη. (B.29)

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APPENDIX B. INTERATOMIC POTENTIALS AND SCATTERING PROCESSES 124

The scattering length then becomes

a =mg

[

1 −mg∫ d3p

(2π)3

1

p2

]

=mg

[

1 − mg

2π2

∫ ∞

0dp]

. (B.30)

We have taken the limit η → 0. The integral is divergence and this is because we havereplaced the true potential by a singular delta-function potential. Note that true potentialalways goes to zero in the limit p → ∞, see Fig. B.2, and so the integral always converges.This is the price we have to pay by using a simple potential. However, we have systematicways of dealing with these divergences and the procedure is called renormalization.

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